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Page No 28.13:

Question 1:

Find the vector equation of a plane passing through a point with position vector 2i^-j^+k^ and perpendicular to the vector 4i^+2j^-3k^.

Answer:

We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a = 2 i^ - j^ + k^ and n = 4 i^ + 2 j^ - 3 k^, we get r. i^ + 2 j^ - 3 k^ = i^ - j^ + k^. i^ + 2 j ^- 3 k^r. i^ + 2 j^ - 3 k^ = 8 - 2 - 3r. i^ + 2 j ^- 3 k ^= 3 

Page No 28.13:

Question 2:

Find the Cartesian form of the equation of a plane whose vector equation is
(i) r·12i^-3j^+4k^+5=0

(ii) r·-i^+j^+2k^=9

Answer:

i Substituting r = x i ^+ y j^ + z k^ in the given equation, we getx i ^+ y j^ + z k^. 12 i^- 3 j^ + 4 k ^ + 5 = 012x - 3y + 4z + 5 = 0

ii Substituting r = x i^ + y j^ + z k^ in the given equation, we getx i ^+ y j^ + z k^ . - i^ + j ^+ 2 k^ = 9-x + y + 2z = 9

Page No 28.13:

Question 3:

Find the vector equations of the coordinate planes.

Answer:

Vector equation of XY-planeThis plane is passing through the origin whose position vector is a =0  and perpendicular to z-axis whose position vector is k^.So, the equation of the XY-plane isr. n=a. nr. k^=0. k^r. k^=0Vector equation of YZ-planeThis plane is passing through the origin whose position vector is a =0  and perpendicular to x-axis whose position vector is i^.So, the equation of the YZ-plane isr. n =a. nr. i^ = 0. i^r. i ^= 0Vector equation of XZ-planeThis plane is passing through the origin whose position vector is a = 0 and perpendicular to y-axis whose position vector is j^.So, the equation of the XZ-plane isr. n=a. nr. j^=0. j^r. j^=0

Page No 28.13:

Question 4:

Find the vector equation of each one of following planes.
(i) 2xy + 2z = 8
(ii) x + yz = 5
(iii) x + y = 3

Answer:

i The given equation of plane is2x-y+2z=8xi^+yj^+zk^. 2 i^-j^+2k^=8r. 2 i^-j^+2k^=8, which is the vector equation of the plane.(Because the vector equation of the plane is rn=an)

ii The given equation of plane isx+y-z=8xi^+yj^+zk^. i^+j^-k^=8r. i^+j^-k^=8, which is the vector equation of the plane.(Because the vector equation of the plane is rn=an)

iii The given equation of plane isx+y=3xi^+yj^+zk^. i^+j^+0k^=3xi^+yj^+zk^. i^+j^=3r.  i^+j^=3, which is the vector equation of the plane.(Because the vector equation of the plane is rn=an)

Page No 28.13:

Question 5:

Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).

Answer:

Since the given plane passes through the point (1, -1, 1) and is normal to the line joining   (1, 2, 5) and (-1, 3, 1), n = AB = OB - OA = -i^ + 3j^ + k^ - i^ + 2j^ + 5k^ = -2 i ^+ j^ - 4k^We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a =  i^ - j^ + k^ and n  =  - 2 i^ +  j ^- 4 k^, we get  r. -2 i^ + j^ - 4k^ =  i^  -  j^  +  k^ . - 2 i ^+ j ^- 4k^r. -2 i^ + j^ - 4k^ = -2 - 1 - 4r. -2 i ^- j^ + 4k^ = -7r. 2 i ^- j^ + 4k^ = 7For Cartesian form, we need to substitute r= xi^ + yj^ + zk^ in the vector equation.Then, we getxi^+yj^+zk^. 2 i^-j^+4k^=72x-y+4z=7

Page No 28.13:

Question 6:

 n is a vector of magnitude 3 and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to n .

Answer:

Let α, β and γ be the angles made by n with x, y and z-axes respectively.Given thatα = β = γcos α = cos β = cos γl=m=n, where l,m, n are direction cosines of n.But l2 + m2 + n2 = 1 l2 + l2 + l2 = 1 3 l2 = 1 l2 = 13 l = 13  (Since α is acute, l = cos α >0)Thus, n= 3 13i ^+ 13j ^+ 13k^ = i^ + j ^+ k^ (Using r =  r l i^ + m j^ + n k^)We know that the vector equation of the plane passing through a point a and normal to n isr. n = a. nSubstituting a =  2 i^  +  j^  -  k^ and n  =  i^ + j^ + k^, we get r. i^ + j^ + k^ =  2 i^ + j^ - k^. i^ + j^ + k^r. i^ + j^ + k^ = 2 + 1 - 1r. i ^+ j^ + k^ = 2For the Cartesian form, we need to substitute r = xi ^+ yj ^+ zk^ in the vector equation.Then, we getxi ^+ yj ^+ zk^. i ^+ j^ + k^ = 2x + y + z = 2

Page No 28.13:

Question 7:

The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.

Answer:

The normal is passing through the points (0, 0, 0) and (12, -4, 3). So,n = AB = OB - OA  =  12 i ^- 4 j ^+ 3 k^ - 0 i ^+ 0 j ^+ 0 k^ = 12 i ^- 4 j ^+ 3 k^Since the plane passes through (12, -4,  3), a  =  12 i^ - 4 j ^+ 3 k^We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a =  i^  -  j^  +  k^ and n =  4 i^  +  2 j^  -  3 k^, we get r. 12 i^ - 4 j^ + 3 k^ = 12 i^ - 4 j^ + 3 k^. 12 i^ - 4 j ^+ 3 k^r. 12 i^ - 4 j ^+ 3 k^ = 144 + 16 + 9r. 12 i ^- 4 j^ + 3 k^ = 169r. 12 i^ - 4 j ^+ 3 k^ = 169Substituting r = xi^ + yj^ + zk^ in the vector equation, we getxi^ + yj ^+ zk^. 12 i ^- 4 j^ + 3 k^ = 169 12x - 4y + 3z = 169

Page No 28.13:

Question 8:

Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.

Answer:

We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a  =  2 i^ + 3 j^ + k^ and n = 5 i^ + 3 j^ + 2 k^, we get r. i^ + 3 j^+ 2 k^ = i^ + 3 j^ + k^. i^ + 3 j^+ 2 k^r. i^ + 3 j^+ 2 k^ = 10 + 9 + 2r. i^ + 3 j^+ 2 k^ = 21Substituting r = xi^ + yj ^+ zk^ in the vector equation, we getxi ^+ yj^ + zk^. i^ + 3 j^ + 2 k^ = 215x + 3y + 2z = 21

Page No 28.13:

Question 9:

If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.

Answer:

The normal is passing through the points (0, 0, 0) and (2, 3 ,-1). So,n = OP = 2 i^+3 j^- k^-0 i^+0 j^+0 k^=2 i^+3 j^- k^Since the plane passes through the point (2, 3 ,-1), a =  2 i^ + 3 j ^- k^We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a =  2 i^ + 3 j^ - k^ and n = 2 i^ + 3 j^ - k^, we get  r. 2 i^+3 j^- k^=2 i^+3 j^- k^. 2 i^+3 j^- k^r. 2 i^+3 j^- k^=4+9+1r. 2 i^+3 j^- k^=14r. 2 i^+3 j^- k^=14Substituting r=xi^+yj^+zk^ in the vector equation, we getxi^+yj^+zk^. 2 i^+3 j^- k^=142x+3y-z=14

Page No 28.13:

Question 10:

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.

Answer:

The given equation of the plane is2x + y - 2z = 3Dividng both sides by 3, we get2x3 + y3 + -2z3 = 33x32 + y3 + z-32 = 1 ... 1We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c isxa + yb + zc = 1 ... 2Comparing (1) and (2), we geta=32; b=3; c=-32Finding the direction cosines of the normalThe given equation of the plane is2x + y - 2z = 8xi^ + yj^ + zk^. 2 i^ + j^ - 2k^ = 8r. 2 i^ + j^ - 2k^ = 8, which is the vector equation of the plane.(Because the vector equation of the plane is rn=an,where the normal to the plane, n=2 i^+j^-2k^.)n=4+1+4=3So, the unit vector perpendicular to n = nn=2 i^+j^-2k^323i^+13j^-23k^So, the direction cosines of the normal to the plane are, 23, 13, -23.

Page No 28.13:

Question 11:

A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point 3i^+j^-k^. Find the vector and Cartesian forms of the equation of the plane.

Answer:

The normal is passing through the points (0, 0, 0) and (3, 1, -1). So,n = OP = 3 i^+ j^- k^-0 i^+0 j^+0 k^ = 3 i^+ j^- k^Since the plane passes through (1, -2, 5), a  =  i^-2 j^+5 k^We know that the vector equation of the plane passing through a point a and normal to n isr. n = a. nSubstituting a =  i^ - j^ + k^ and n  =  4 i^ + 2 j ^- 3 k^, we get r. 3 i^ + j^- k^=i^-2 j^+ 5 k^. 3 i^+ j^- k^r. 3 i^+ j^- k^=3-2-5r. 3 i^+ j^- k^=-4r. 3 i^+ j^- k^=-4Substituting r=xi^+yj^+zk^ in the vector equation, we getxi^ + yj^ + zk^. 3 i^ + j^- k^ = -43x + y - z = -4

Page No 28.13:

Question 12:

Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

Answer:

The normal is passing through the points (1, 2, 3) and (3, 4, 5). So,n = AB = OB - OA = 3 i^+4 j^+5 k^-i^+2 j^+3 k^ = 2 i^+2 j^+2 k^Mid-point of AB 1+32, 2+42, 3+52 = 2, 3, 4Since the plane passes through 2, 3, 4a =  2 i^+3 j^+4 k^We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a =  i^ - j^ + k^ and n = 4 i^ + 2 j^ - 3 k^, we get  r. 2 i^+2 j^+2 k^ = 2 i^+3 j^+4 k^. 2 i^+2 j^+2 k^r. 2 i^+2 j^+2 k^ = 4+6+8r. 2 i^+2 j^+2 k^ = 18r. 2 i^+j^+k^ = 18r. i^+j^+k^ = 9Substituting r = xi^+yj^+zk^ in the vector equation, we getxi^+yj^+zk^. i^+j^+k^ = 9x+y+z = 9

Page No 28.13:

Question 13:

Show that the normals to the following pairs of planes are perpendicular to each other.

(i) xy + z − 2 = 0 and 3x + 2yz + 4 = 0

(ii) r·2i^-j^+3k^=5 and r·2i^-2j^-2k^=5

Answer:

i Let n1 and n2 be the vectors which are normals to the planes x-y+z=2 and 3x+2y-z=-4 respectively.The given equations of the planes arex - y + z = 2;  3x + 2y - z = -4xi^ + yj^ + zk^. i^ - j^ + k^ = 8; xi^ + yj^ + zk^. 3 i^ + 2 j ^- k^ = -4n1 = i ^- j^ + k^;  n2 = 3 i^ + 2 j^ - k^Now, n1 n2=i^ - j ^+ k^. 3 i^ + 2 j ^- k^ = 3 - 2 - 1 = 0So, the normals to the given planes are perpendicular to each other.

ii Let n1 and n2 be the vectors which are normals to the planes r2 i^-j^+3 k^ = 5 and r2 i^-2 j^-2 k^ = 5 respectively.The given equations of the planes arer2 i^-j^+3 k^ = 5 ; r2 i^-2 j^-2 k^ = 5n1=2 i^-j^+3 k^; n2=2 i^-2 j^-2 k^Now, n1 n2=2 i^-j^+3 k^. 2 i^-2 j^-2 k^=4+2-6=0So, the normals to the given planes are perpendicular to each other.

Page No 28.13:

Question 14:

Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.

Answer:

The given equation of the plane is2x+2y+2z=8xi^+yj^+zk^. 2 i^+2 j^+2k^=8r. 2 i^+2 j^+2k^=8, which is the vector equation of the plane.(Because the vector equation of the plane is rn=an,where the normal to the plane, n=2 i^+2 j^+2k^.)n=4+4+4=2 3So, the unit vector perpendicular to n = nn=2 i^+2 j^+2k^2 313i^+13j^+13k^So, the direction cosines of the normal to the plane are l=13, m=13, n=13Let α, β and γ be the angles made by the given plane with the coordinate axes.Then,l=cos α=13; m=cos β=13; n=cos γ=13cos α=cos β=cos γα=β=γSo, the given plane is equally inclined to the coordinate axes.



Page No 28.14:

Question 15:

Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1.

Answer:

The given equation of the plane is12x-3y+4z=1xi^+yj^+zk^. 12 i^-3j^+4k^=1r. 12 i^-3j^+4k^=1, which is the vector equation of the plane.(Because the vector equation of the plane is rn=an)So, the normal vector, n=12 i^-3j^+4k^n=144+9+16=13Unit vector parallel to n=nn=12 i^-3j^+4k^13So, the vector of magnitude 26 units normal to the plane=26 ×12 i^-3j^+4k^13=2 12 i^-3j^+4k^=24 i^-6j^+8k^

Page No 28.14:

Question 16:

If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.

Answer:

The normal is passing through the points (4, -1, 2) and (-10, 5, 4). So,n = AB = OB - OA = -10 i^+5  j^+4 k^-4 i^- j^+2 k^=-14 i^+6 j^+2 k^Since the plane passes through (-10, 5, 4), a = -10 i^+5  j^+4 k^We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a =-10i^+5  j^+4 k^ and n =-14 i^+6 j^+2 k^, we get  r. -14 i^+6 j^+2 k^=-10i^+5  j^+4 k^. -14 i^+6 j^+2 k^r. -14 i^+6 j^+2 k^=140+30+8r. -2 7 i^-3 j^-k^=178r. 7 i^-3 j^-k^=-89Substituting r=xi^+yj^+zk^ in the vector equation, we getxi^+yj^+zk^. 7 i^-3 j^-k^=-897x-3y-z=-897x-3y-z+89=0

Page No 28.14:

Question 17:

Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.

Answer:

The normal is passing through the points (-1, 2, 3) and (3, -5, 6). So,n = AB = OB - OA = 3 i^-5 j^+6 k^--i^+2 j^+3 k^=4 i^-7 j^+3 k^Mid-point of AB -1+32, 2-52, 3+62=1, -3 2, 92Since the plane passes through 1, -3 2, 92a =   i^-32 j^+92 k^We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a =  i^ - j^ + k^ and n = 4 i^-7 j^+3 k^, we get  r. 4 i^-7 j^+3 k^ = i^-32 j^+92 k^. 4 i^-7 j^+3 k^r. 4 i^-7 j^+3 k^=28Substituting r=xi^+yj^+zk^ in the vector equation, we getxi^+yj^+zk^. 4 i^-7 j^+3 k^ = 284x-7y+3z = 284x-7y+3x-28 = 0

Page No 28.14:

Question 18:

Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.

Answer:

We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a = 5 i^ +2 j^ -4 k^ and n = 2 i^ +3 j^k^ (because the direction ratios of n are 2, 3, -1), we getr. i^ +3 j^k^= 5 i^ +2 j^ -4 k^. i^ +3 j^k^r. i^ +3 j^k^=10+6+4r. i^ +3 j^k^=20For Cartesian form, we need to substitute r = x i^+y j^+z k^ in this equation. Then, we getx i^+y j^+z k^. i^ +3 j^k^=202x+3y-z=20

Page No 28.14:

Question 19:

If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP.

Answer:

The normal is passing through the points (0, 0, 0) and (1, 2, -3). So,n = OP = i^+2 j^-3 k^-0 i^+0 j^+0 k^=i^+2 j^-3 k^Since the plane passes through (1, 2, -3), a =   i^+2 j^-3 k^We know that the vector equation of the plane passing through a point a and normal to n isr. n=a. nSubstituting a =  2 i^+3 j^- k^ and n = i^+2 j^-3 k^ in the relation, we get  r. i^+2 j^-3 k^=i^+2 j^-3 k^. i^+2 j^-3 k^r. i^+2 j^-3 k^=1+4+9r. i^+2 j^-3 k^=14r. i^+2 j^-3 k^=14Substituting r=xi^+yj^+zk^ in the vector equation, we getxi^+yj^+zk^. i^+2 j^-3 k^=14x+2y-3z=14

Page No 28.14:

Question 20:

If O is the origin and the coordinates of A are (abc). Find the direction cosines of OA and the equation of the plane through A at right angles to OA.              [NCERT EXEMPLAR]

Answer:


It is given that O is the origin and the coordinates of A are (abc).

The direction ratios of OA are proportional to 

a-0,b-0,c-0 or a, b, c

∴ Direction cosines of OA are

aa2+b2+c2,ba2+b2+c2,ca2+b2+c2

The normal vector to the required plane is ai^+bj^+ck^.

The vector equation of the plane through A(abc) and perpendicular to OA is

r-ai^+bj^+ck^.ai^+bj^+ck^=0                                        r-a.n=0r.ai^+bj^+ck^=ai^+bj^+ck^.ai^+bj^+ck^r.ai^+bj^+ck^=a2+b2+c2

The Cartesian equation of this plane is

xi^+yj^+zk^.ai^+bj^+ck^=a2+b2+c2Or ax+by+cz=a2+b2+c2

Page No 28.14:

Question 21:

Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.

Answer:


The equation of the plane in the intercept form is xa+yb+zc=1, where ab and are the intercepts on the xy and z-axis, respectively.

It is given that the intercepts made by the plane on the xy and z-axis are 3, –4 and 2, respectively.

∴ a = 3, b = −4, c = 2

Thus, the equation of the plane is 

x3+y-4+z2=14x-3y+6z=12
xi^+yj^+zk^.4i^-3j^+6k^=12r.4i^-3j^+6k^=12
This is the vector form of the equation of the given plane.



Page No 28.19:

Question 1:

Find the vector equation of a plane which is at a distance of 3 units from the origin and has k^ as the unit vector normal to it.

Answer:

Given thatnormal vector, n=i^Now, n^=nn=k^k^=k^1=k^The equation of a plane in normal form isr. n^=d  (where is the distance of the plane from the origin)Substituting n^=k^ and = 3 in the relation, we get r. k^=3

Page No 28.19:

Question 2:

Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector i^-2j^-2k^.

Answer:

It is given that the normal vector, n=i^-2 j^-2 k^Now, n^=nn=i^-2 j^-2 k^1+4+4=i^-2 j^-2 k^3=13 i^-23 j^-23k^The equation of a plane in normal form isr. n^=d (where is the distance of the plane from the origin)Substituting n^=13 i^-23 j^-23k^ and = 5Here,r. 13 i^-23 j^-23k^=5

Page No 28.19:

Question 3:

Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

Answer:

The given equation of the plane is2x - 3y - 6z = 14 ... 1Now, 22+-32+-62 4+9+36 = 49 = 7Dividing (1) by 7, we get27x - 37y - 67z = 2 ... 2The Cartesian equation of the normal form of a plane islx+ my + nz = p... 3,where l, m and n are direction cosines of normal to the plane and p is the length of the perpendicular from the origin to the plane.Comparing (1) and (2), we getdirection cosines: l=27, m=-37, n=-6 7 andlength of the perpendicular from the origin to the plane: p=2

Page No 28.19:

Question 4:

Reduce the equation r·i^-2j^+2k^+6=0 to normal form and, hence, find the length of the perpendicular from the origin to the plane.

Answer:

The given equation of the plane isr. i^-2 j^+2 k^+6 = 0r. i^-2 j^+2 k^ =-6 or r. n = -6, where n = i^-2 j^+2 k^n =1+4+4 = 3For reducing the given equation to normal form, we need to divide it by n. Then, we getr. nn = -6nr. i^-2 j^+2 k^3 = -63r. 13 i^-23 j^+23k^ = -2Dividing both sides by -1, we getr. -13 i^+23 j^-23k^ = 2 ... 1The equation of the plane in normal form isr. n ^= d ... 2(where is the distance of the plane from the origin)Comparing (1) and (2), length of the perpendicular from the origin to the plane = d = 2 units

Page No 28.19:

Question 5:

Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.

Answer:

The given equation of the plane is2x-3y+6z+14 = 02x-3y+6z =-14 ... 1Now, 22 + -32 + 62 4 + 9 + 36  = 49  =  7Dividing (1) by 7, we get27x - 37y + 67z = -2 Multiplying both sides by -1, we get-27x+37y-67z=2This is the normal form of the given equation of the plane.

Page No 28.19:

Question 6:

The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane.

Answer:

It is given that the direction ratios of the normal vector n are 12, -3, 4.So, n=12 i^-3 j^+4 k^n=122+-32+42=144+9+16=169=13Now, n^=nn=12 i^-3 j^+4 k^13=1213 i^-313 j^+413 k^Length of the perpendicular from the origin to the plane, = 5Equation of the plane in normal form isr. n^=dr. 1213 i^-313 j^+413 k^=5

Page No 28.19:

Question 7:

Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.

Answer:

The given equation of the plane isx+2y+3z-6=0x+2y+3z=6 r. i^ + 2 j ^+ 3 k^ = 6 or  rn   =  6,where n = i^+2 j^+3 k^... 1Now, n = 12 + 22 + 32=1 + 4 + 9 14Unit vector to the plane, n = nn  =  i ^+ 2 j^ + 3 k^14= 114i^ + 214j^ +314k^

Page No 28.19:

Question 8:

Find the equation of a plane which is at a distance of 33 units from the origin and the normal to which is equally inclined to the coordinate axes.

Answer:

Let α, β and γ be the angles made by n with x, y and z-axes, respectively.It is given thatα=β=γcos α=cos β=cos γl=m=n, where l,m, n are direction cosines of n.But l2+m2+n2=1l2+l2+l2=13 l2=1l2=13l=13So, l=m=n=13It is given that the length of the perpendicular of the plane from the origin, = 3 3The normal form of the plane is lx+my+nz=p13x + 13y +13z = 33x + y + z = 33 3 x + y + z = 9

Page No 28.19:

Question 9:

Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane.

Answer:

The normal is passing through the points (1, 4, 2) and B (2, 3, 5).So, n = AB = OB - OA = 2 i^+3 j^+5 k^-i^+4 j^+2 k^= i^-j^+3 k^We know that the vector equation of the plane passing through a point (1, 2, 1) (a) and normal to n isr. n=a. nSubstituting a =  i^ +2 j^ + k^ and n = i^-j^+3 k^, we get  r. i^-j^+3 k^= i^ +2 j^ + k^. i^-j^+3 k^r. i^-j^+3 k^=1-2+3r. i^-j^+3 k^=2 ... 1To find the perpendicular distance of this plane from the origin, we have to reduce this to normal form.We have n=i^-j^+3 k^; n=1+1+9=11Dividing (1) by 11, we getr. 111i^-111j^+311 k^ = 211, which is the normal form of plane (1).So, the perpendicular distance of plane (1) from the origin = 211

Page No 28.19:

Question 10:

Find the vector equation of the plane which is at a distance of 629 from the origin and its normal vector from the origin is 2i^-3j^+4k^. Also, find its Cartesian form.

Answer:

Given, normal vector, n=2 i^-3 j^+4 k^Now, n^=nn=i^-3 j^+4 k^4+9+16=i^-3 j^+4 k^29=229 i^-329 j^+429k^The equation of the plane in normal form isr. n^=d (where is the distance of the plane from the origin)Substituting, n^=229 i^-329 j^+429k^ and 629 here, we get r. 229 i^-329 j^+429k^=629... (1)Cartesian formFor Cartesian form, substituting rx i^+y j^+z k^ in (1), we getx i^+y j^+z k^ . 229 i^-329 j^+429k^=  6292x-3y+4z29=6292x-3y+4z=6

Page No 28.19:

Question 11:

Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.

Answer:

The given equation of the plane is2x-3y+4z=6 ... 1Now, 22 + -32 + 42  =  4+9+16 29Dividing (1) by 29, we get229x-329y+429z = 629, which is the normal form of plane (1).So, the length of the perpendicular from the origin to the plane=629



Page No 28.22:

Question 1:

Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).

Answer:



Let A (1, 1, 1), B (1, -1, 1) and C (-7, -3, -5) be the coordinates.The required plane passes through the point  A (1, 1, 1) whose position vector is a = i^ + j^ + k^ and is normal to the vector n given byn = AB × AC .Clearly, AB = OB - OA =  i ^- j^ + k^ - i ^+ j ^+ k^ = 0 i^ - 2 j^ + 0 k^AC = OC - OA =-7 i^ - 3 j^ - 5 k^ - i ^+ j^ + k^ = -8 i^ -4 j^ -6 k^n = AB × AC =i^j^k^0-20-8-4-6 = 12 i^ + 0 j^ -16 k^The vector equation of the required plane isr. n = a. nr. 12 i^ + 0 j^ -16 k^ = i^ + j^ + k^. 12 i^ + 0 j^ -16 k^r. 4 3 i^ - 4 k^ = 12 + 0 - 16r. 4 3 i^ - 4 k^ = -4r. 3 i^ - 4 k^ = -1r. 3 i ^- 4 k^ + 1 = 0



Page No 28.23:

Question 2:

Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).

Answer:



The required plane passes through the point  P (2, 5, -3) whose position vector is a=2 i^+5 j^-3 k^ and is normal to the vector n given byn = PQ×PR.Clearly, PQ = OQ - OP = -2 i^ - 3 j^ + 5 k^ - i^ + 5 j^ - 3 k^ = -4 i ^- 8 j^ + 8 k^PR = OR - OP = 5 i^ + 3 j^ - 3 k^ - i^ + 5 j ^- 3 k^ = 3 i ^- 2 j^ - 0 k^n = PQ × PR = i^j^k^-4-883-20 = 16 i^ + 24 j^ + 32 k^The vector equation of the required plane isr. n = a. nr. 16 i^+24 j^+32 k^ = i^+5 j^-3 k^. 16 i^+24 j^+32 k^r. 8 2 i^+3 j^+4 k^ = 32+120-96r. 8 2 i^+3 j^+4 k^ = 56r. 2 i^+3 j^+4 k^ = 7

Page No 28.23:

Question 3:

Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C (0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that 1p2=1a2+1b2+1c2.

Answer:



The required plane passes through the point  A (a, 0, 0) whose position vector is a=i^ + 0 j ^+ 0 k^ and is normal to the vector n given byn=AB × AC.Clearly, AB = OB - OA = 0 i ^+ b j^ + 0 k^ - i^ + 0 j^ + 0 k^ = -a i ^+ b j^ + 0 k^AC = OC - OA = 0 i^ + 0 j^ + c k^ - i ^+ 0 j^ + 0 k^ = -a i^ + 0 j^ + c k^n = AB × AC = i^j^k^-a b 0-a0c = bc i^ + ac j^ + ab k^The vector equation of the required plane isr. n = a. nr. bc i^ + ac j^  + ab k^ = i^ + 0 j^ + 0 k^. bc i^ + ac j^ +ab k^r. bc i^ + ac j^ + ab k^ = abc + 0 + 0r. bc i^ + ac j^  + ab k^ = abc ... 1Now, n = bc2 + ac2 + ab2  =  b2c2 + a2c2 + a2b2For reducing (1) to normal form, we need to divide both sides of (1) by b2c2+a2c2+a2b2. Then, we getr. bc i^+ac j^ +ab k^b2c2+a2c2+a2b2 = abc b2c2+a2c2+a2b2, which is the normal form of plane (1).So, the distance of plane (1) from the origin,p = abc b2c2 + a2c2 + a2b2,1p =  b2c2 + a2c2 + a2b2abc1p2  = b2c2 + a2c2 + a2b2a2b2c21p2 = 1a2 + 1b2 + 1c2

Page No 28.23:

Question 4:

Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).

Answer:



Let A (1, 1, -1), B (6, 4, -5) and C (-4, -2, 3).The required plane passes through the point  A (1, 1, -1) whose position vector is a = i ^+ j ^- k^ and is normal to the vector n given byn=AB × ACClearly, AB = OB - OA  = 6 i^ + 4 j^ - 5 k^ - i^ + j ^- k^ = 5 i^ + 3 j^ - 4 k^AC = OC - OA = -4 i^ - 2 j^ + 3 k^ - i^ + j ^- k^ = -5 i^ - 3 j^ + 4 k^n=AB × AC = i^j^k^53-4-5-34 = 0 i^ + 0 j^ +0 k^ = 0So, the given points are collinear.Thus, there will be infinite number of planes passing through these points.Their equations (passing through (1, 1, -1) are given bya x - 1 + b y - 1 + c z + 1 = 0 ... 1Since this passes through B (6, 4, -5),a 6 - 1 + b 4 - 1 + c -5 + 1 = 05a + 3b - 4c = 0 ... 2From (1) and (2), the equations of the infinite planes area x - 1 + b y - 1 + c z + 1 = 0, where 5a + 3b - 4c = 0.

Page No 28.23:

Question 5:

Find the vector equation of the plane passing through the points 3i^+4j^+2k^, 2i^-2j^-k^ and 7i^+6k^.

Answer:


Let A (3, 4, 2), B (2, -2, -1) and C (7, 0, 6) be the points represented by the given position vectors.The required plane passes through the point  A (3, 4, 2) whose position vector is a=3 i^+4 j^+2 k^ and is normal to the vector n given byn=AB×AC.Clearly, AB=OB-OA=2 i^-2 j^-k^-i^+4 j^+2 k^=- i^-6 j^-3 k^AC=OC-OA=7 i^+0  j^+6 k^-i^+4 j^+2 k^=4 i^-4 j^+4 k^n=AB×AC=i^j^k^-1-6-34-44=-36 i^-8 j^ +28 k^The vector equation of the required plane isr. n=a. nr. -36 i^-8 j^ +28 k^=i^+4 j^+2 k^. -36 i^-8 j^ +28 k^r. -4 9 i^+2 j^+7 k^=-108-32+56r. -4 9 i^+2 j^+7 k^=-84r. 9 i^+2 j^+7 k^=21



Page No 28.29:

Question 1:

Find the angle between the given planes.
(i) r·2i^-3j^+4k^=1 and r·-i^+j^=4

(ii) r·2i^-j^+2k^=6 and r·3i^+6j^-2k^=9

(iii) r·2i^+3j^-6k^=5 and r·i^-2j^+2k^=9

Answer:

i We know that the angle between the planes r.n1 = d1, r. n2=d2 is given bycos θ=n1. n2n1 n2Here, n1=2 i^-3 j^+4 k^; n2=- i^+j^+0 k^So, cos θ=2 i^-3 j^+4 k^. - i^+j^+0 k^2 i^-3 j^+4 k^ - i^+j^+0 k^ = -2-34+9+16 1+1+0 = -529 2 = -558θ=cos-1-558

ii We know that the angle between the planes r.n1 = d1, r. n2=d2 is given bycos θ=n1. n2n1 n2Here, n1=2 i^- j^+2 k^; n2=3 i^+6 j^-2 k^So, cos θ=2 i^- j^+2 k^. 3 i^+6 j^-2 k^2 i^- j^+2 k^ 3 i^+6 j^-2 k^=6-6-44+1+4 9+36+4=-43 7=-421θ=cos-1-421

iii We know that the angle between the planes r.n1 = d1, r. n2=d2 is given bycos θ=n1. n2n1 n2Here, n1=2 i^ +3 j^-6 k^; n2= i^-2 j^+2 k^So, cos θ=2 i^ +3 j^-6 k^.  i^-2 j^+2 k^2 i^ +3 j^-6 k^  i^-2 j^+2 k^=2-6-124+9+36 1+4+4=-167 3=-1621θ=cos-1-1621

Page No 28.29:

Question 2:

Find the angle between the planes.
(i) 2xy + z = 4 and x + y + 2z = 3
(ii) x + y − 2z = 3 and 2x − 2y + z = 5
(iii) xy + z = 5 and x + 2y + z = 9
(iv) 2x − 3y + 4z = 1 and − x + y = 4
(v) 2x + y − 2z = 5 and 3x − 6y − 2z = 7

Answer:

i We know that the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given bycos θ=a1a2+b1b2+c1c2a12+b12+c12 a22+b22+c22So, the angle between 2x - y + z = 4 and x + y + 2z = 3 is given by cos θ=2 1 + -1 1 + 1 222 + -12 + 12 12 + 12 + 22=2 - 1 + 24 + 1 + 1 1 + 1 + 4=36 6=36=12θ = cos-112 = π3

ii We know that the angle between the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 is given bycos θ=a1a2+b1b2+c1c2a12+b12+c12 a22+b22+c22So, the angle between x+y-2z=3 and 2x-2y+z=5 is given by cos θ=1 2+1 -2+-2 112+12+-22 22+-22+12=2-2-21+1+4 4+4+1=-26 9=-236θ=cos-1-236
iii We know that the angle between the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 is given bycos θ=a1a2+b1b2+c1c2a12+b12+c12 a22+b22+c22So, the angle between x-y+z=5 and x+2y+z=9 is given by cos θ=1 1+-1 2+1 112+-12+12 12+22+12=1-2+11+1+1 1+4+1=03 6=0θ=cos-10=π2

iv We know that the angle between the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 is given bycos θ=a1a2+b1b2+c1c2a12+b12+c12 a22+b22+c22So, the angle between 2x-3y+4z=1 and -x+y+0z=4 is given by cos θ=2 -1+-3 1+4 022+-32+42 -12+12+02=-2-3+04+9+16 1+1+0=-529 2=-558θ=cos-1-558

v We know that the angle between the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 is given bycos θ=a1a2+b1b2+c1c2a12+b12+c12 a22+b22+c22So, the angle between 2x+y-2z=5 and 3x-6y-2z=7 is given by cos θ=2 3+1 -6+-2 -222+12+-22 32+-62+-22=6-6+44+1+4 9+36+4=43 7=421θ=cos-1421

Page No 28.29:

Question 3:

Show that the following planes are at right angles.
(i) r·2i^-j^+k^=5 and r·-i^-j^+k^=3

(ii) x − 2y + 4z = 10 and 18x + 17y + 4z = 49

Answer:

i We know that the planes r.n1 = d1, r. n2=d2 are perpendicular to each other only if n1n2=0.Here, n1=2 i^- j^+ k^;  n2=- i^-j^+k^Now, n1. n2 = 2 i^- j^+ k^. - i^-j^+k^=-2 + 1 + 1 = 0So,the given planes are perpendicular.

ii We know that the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 are perperndicular to each other only if a1a2+b1b2+c1c2=0.The given planes are x-2y+4z=10 and 18x+17y+4z=49.a1=1; b1=-2; c1=4; a2=18; b2=17; c2=4Now, a1a2+b1b2+c1c2=1 18+-2 17+4 4=18-34+16=0So, the given planes are perpendicular.

Page No 28.29:

Question 4:

Determine the value of λ for which the following planes are perpendicular to each other.
(i) r·i^+2j^+3k^=7 and r·λi^+2j^-7k^=26

(ii) 2x − 4y + 3z = 5 and x + 2y + λz = 5

(iii) 3x − 6y − 2z = 7 and 2x + y − λz = 5

Answer:

i We know that the planes r.n1 = d1, r. n2=d2 are perpendicular to each other only if n1n2=0.Here, n1= i^ + 2 j ^+ 3 k^;  n2 = λ i^ + 2 j ^- 7 k^The given planes are perpendicular.n1. n2 = 0i^+2 j^+3 k^. λ i^+2 j^-7 k^ = 0λ+4-21=0λ-17=0λ=17

ii We know that the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 are perperndicular to each other only if a1a2+b1b2+c1c2=0The given planes are 2x-4y+3z=5 and x+2y+λz=5.a1=2; b1=-4; c1=3; a2=1; b2=2; c2=λIt is given that the given planes are perpendicular. a1a2+b1b2+c1c2=02 1+-4 2+3 λ=02-8+3λ=03λ=6λ=2

iii We know that the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 are perperndicular to each other only if a1a2+b1b2+c1c2=0The given planes are 3x-6y-2z=7 and 2x+y-λz=5.a1=3; b1=-6; c1=-2; a2=2; b2=1; c2=-λThe given planes are perpendicular. a1a2+b1b2+c1c2=03 2+-6 1+-2 -λ=06-6+2λ=02λ=0λ=0

Page No 28.29:

Question 5:

Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.

Answer:

The equation of any plane passing through (-1, -1, 2) isa x+1+b y+1+c z-2=0 ... 1It is given that (1) is perpendicular to each of the planes 3x+2y-3z=1 and 5x-4y+z=5. Then,3a+2b-3c=0 ... 25a-4b+c=0 ... 3Solving (1), (2) and (3), we getx+1y+1z-232-35-41 = 0-10 x + 1 - 18 y + 1 -22 z - 2 = 05 x + 1 + 9 y + 1 + 11 z - 2 = 05x + 5 + 9y + 9 + 11z - 22 = 05x + 9y + 11z - 8 = 0

Page No 28.29:

Question 6:

Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

Answer:

The equation of any plane passing through (1, -3, -2) isa x-1+b y+3+c z+2=0 ... 1It is given that (1) is perpendicular to the planes x+2y+2z=5 and 3x+3y+2z=8. Then,a+2b+2c=0 ... 23a+3b+2c=0 ... 3Solving (1), (2) and (3), we getx-1y+3z+2122332=0-2 x-1+4 y+3-3 z+2=0-2x+2+4y+12-3z-6=02x-4y+3z-8=0

Page No 28.29:

Question 7:

Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2yz = 1 and 3x − 4y + z = 5.

Answer:

The equation of any plane passing through the origin (0, 0, 0) isa x-0+b y-0+c z-0=0 ax+by+cz=0... 1It is given that (1) is perpendicular to the planes x+2y-z=1 and 3x-4y+z=5. Then,a+2b-c=0 ... 23a-4b+c=0 ... 3Solving (1), (2) and (3), we getxyz12-13-41=0-2x-4y-10z=0x+2y+5z=0

Page No 28.29:

Question 8:

Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.

Answer:

The equation of any plane passing through (1, -1, 2) isa x-1+b y+1+c z-2=0 ... 1It is given that (1) is passing through (2, -2, 2). So,a 2-1+b -2+1+c 2-2=0 a-b+0c=0... 2It is given that (1) is perpendicular to the plane 6x-2y+2z=9. So,6a-2b+2c=03a-b+c=0... 3Solving (1), (2) and (3), we getx-1y+1z-21-103-11=0-1 x-1-1 y+1+2 z-2=0-x+1-y-1+2z-4=0x+y-2z+4=0

Page No 28.29:

Question 9:

Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

Answer:

The equation of any plane passing through (2, 2, 1) isa x-2+b y-2+c z-1=0 ... 1It is given that (1) is  passing through (9, 3, 6). So,a 9-2+b 3-2+c 6-1=07a+b+5c=0... 2It is given that (1) is perpendicular to the plane 2x+6y+6z=1. So,2a+6b+6c=0a+3b+3c=0... 3Solving (1), (2) and (3), we getx-2y-2z-1715133=0-12 x-2-16 y-2+20 z-1=03 x-2+4 y-2 -5 z-1=03x+4y-5z=9

Page No 28.29:

Question 10:

Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.

Answer:

The equation of any plane passing through (-1, 1, 1) isa x+1+b y-1+c z-1=0 ... 1It is given that (1) is  passing through (1, -1, 1). So,a 1+1+b -1-1+c 1-1=0 2a-2b+0c=0... 2It is given that (1) is perpendicular to the plane x+2y+2z=5. So,a+2b+2c=0 ... 3Solving (1), (2) and (3), we getx+1y-1z-12-20122=0-4 x+1-4 y-1+6 z-1=02 x+1+2 y-1-3 z-1=02x+2y-3z+3=0

Page No 28.29:

Question 11:

Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.

Answer:

The equation of the plane parallel to the plane ZOX is y=b ... 1, where b is a constant.It is given that this plane passes through (0, 3, 0). So,3=bSubstituting this value in (1), we get y=3, which is the required equation of the plane.

Page No 28.29:

Question 12:

Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.

Answer:

The equation of any plane passing through (1, -1, 2) isa x-1+b y+1+c z-2=0 ... 1It is given that (1) is perpendicular to the plane 2x+3y-2z=5. So,2a+3b-2c=0... 2It is given that (1) is perpendicular to the plane x+2y-3z=8. So,a+2b-3c=0 ... 3Solving (1), (2) and (3), we getx-1y+1z-223-212-3=0-5 x-1+4 y+1+1 z-2=05x-4y-z=7

Page No 28.29:

Question 13:

Find the equation of the plane passing through (a, b, c) and parallel to the plane r·i^+j^+k^=2.

Answer:

Substituting r=x i^+y j^+z k^  in the given equation of the plane, we getx i^+y j^+z k^. i^+j^+k^=2x+y+z-2=0 ... (1)The equation of a plane which is parallel to plane (1) is of the formx+y+z=k... 2It is given that plane (2) is passing through the point a, b, c.  So,a+b+c=kSubstituting this value of k in (2), we get x+y+z=a+b+c, which is the required equation of the plane.

Page No 28.29:

Question 14:

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Answer:

The equation of any plane passing through point (-1, 3, 2) isa x+1+b y-3+c z-2=0 ... 1It is given that (1) is perpendicular to the plane x+2y+3z=5. So,a+2b+3c=0 ... 2It is given that (1) is perpendicular to the plane 3x+3y+z=0. So,3a+3b+c=0 ... 3Solving (1), (2) and (3), we getx+1y-3z-2123331=0-7 x+1+8 y-3-3 z-2=07x-8y+3z+25=0

Page No 28.29:

Question 15:

Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10.

Answer:

The equation of any plane passing through (2, 1, -1) isa x-2+b y-1+c z+1=0 ... 1It is given that (1) is  passing through (-1, 3, 4). So,a -1-2+b 3-1+c 4+1=0-3a+2b+5c... 2It is given that (1) is perpendicular to the plane x-2y+4z=10. So,a-2b+4c=0... 3Solving (1), (2) and (3), we getx-2y-1z+1-3251-24=018 x-2+17 y-1+4 z+1=018x+17y+4z-49=0



Page No 28.33:

Question 1:

Find the vector equations of the following planes in scalar product form r·n=d:
(i) r=2i^-k^+λi^+μi^-2j^-k^

(ii) r=1+s-t t^+2-s j^+3-2s+2t k^

(iii) r=i^+j^+λi^+2j^-k^+μ-i^+j^-2k^

(iv) r=i^-j^+λi^+j^+k^+μ4i^-2j^+3k^

Answer:

i We know that the equation r=a+λb+μc represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=2 i^+0 j^-k^; b=i^; c=i^-2 j^-k^Normal vector, n=b×c=i^j^k^1001-2-1=0 i^+j^-2 k^=j^-2 k^The vector equation of the plane in scalar product form isr. n=a. nr. j^-2 k^=2 i^+0 j^-k^. j^-2 k^r. j^-2 k^=2

ii The given equation of the plane isr=1+s-t i^+2-s j^+3-2s+2t k^r=i^+2 j^+3 k^+s i^-j-2 k^+t - i^+0 j^+2 k^ We know that the equation r=a+sb+t c represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=i^+2 j^+3 k^; b=i^-j-2 k^; c=- i^+0 j^+2 k^Normal vector, n=b×c=i^j^k^1-1-2-102=-2 i^+0 j^-k^=-2 i^- k^The vector equation of the plane in scalar product form isr. n=a. nr. -2 i^- k^=i^+2 j^+3 k^. -2 i^- k^r. -1 2 i^+k^=-2+0-3r. -1 2 i^+k^=-5r. 2 i^+k^=5

iii We know that the equation r=a+λb+μc represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=i^+ j^+0 k^; b=i^+2 j^-k^; c=-i^+ j^-2 k^Normal vector, n=b×c=i^j^k^12-1-11-2=-3 i^+3 j^+3 k^The vector equation of the plane in scalar product form isr. n=a. nr. -3 i^+3 j^+3 k^=i^+ j^+0 k^. -3 i^+3 j^+3 k^r. -3 i^+3 j^+3 k^=-3+3r. 3 -i^+ j^+k^=0r. -i^+ j^+k^=0

iv We know that the equation r=a+λb+μc represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=i^- j^+0 k^; b=i^+j^+k^; c=4 i-2 j^+3 k^Normal vector, n=b×c=i^j^k^1114-23=5 i^+j^-6 k^The vector equation of the plane in scalar product form isr. n=a. nr. 5 i^+j^-6 k^=i^- j^+0 k^. 5 i^+j^-6 k^r. 5 i^+j^-6 k^=5-1+0r. 5 i^+j^-6 k^=4r. 5 i^+j^-6 k^=4

Disclaimer: The answer given for part (iv) of this problem in the text book is incorrect.

Page No 28.33:

Question 2:

Find the Cartesian forms of the equations of the following planes.
(i) r=i^-j^+s-i^+j^+2k^+ti^+2j^+k^

(ii) r=1+s+ti^+2-s+ti^+3-2s+2tk^

Answer:

i r=i^- j^+ 0 k^+s -i^ + j + 2 k^ + t i^ + 2 j^ + k^ We know that the equation r=a + sb + t c represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=i^ - j^ + 0 k^;  b=-i^ + j + 2 k^; c = i^ + 2 j^ + k^Normal vector, n=b×c=i^j^k^-112121=-3 i^ + 3 j^ - 3 k^The vector equation of the plane in scalar product form isr. n=a. nr. -3 i^ + 3 j^ - 3 k^ = i^ -  j ^+ 0 k^. -3 i ^+ 3 j^ - 3 k^r. -3 i^ - j ^+ k^ =-3 - 3 + 0r. -3 i^ - j^ + k^ = -6r. i^ - j^ + k^ = 2For Cartesian form, let us substitute r = x i^ + y j^ + z k^ here. Then, we getx i^ + y j^ + z k^. i^ - j ^+ k^ = 2x - y + z = 2

ii The given equation of the plane isr=1+s+t i^+2-s+t j^+3-2s+2t k^r=i^+2 j^+3 k^+s i^-j-2 k^+t i^+j^+2 k^ We know that the equation r=a+sb+t c represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=i^+2 j^+3 k^; b=i^-j-2 k^; c=i^+j^+2 k^Normal vector, n=b×c=i^j^k^1-1-2112= 0 i^-4 j^+2 k^=-4 j^+2 k^The vector equation of the plane in scalar product form isr. n=a. nr. -4 j^+2 k^=i^+2 j^+3 k^. -4 j^+2 k^r. -2 2 j^-k^=0-8+6r. -2 2 j^-k^=-2r. 2 j^-k^=1For Cartesian form, let us substitute r=x i^+y j^+z k^ here. Then, we getx i^+y j^+z k^. 2 j^-k^=12y-z=1

Page No 28.33:

Question 3:

Find the vector equation of the following planes in non-parametric form.
(i) r=λ-2μ i^+3-μ j^+2λ+μ k^

(ii) r=2i^+2j^-k^+λi^+2j^+3k^+μ5i^-2j^+7k^

Answer:

i The given equation of the plane isr=λ-2μ i^+3-μ j^+2λ+μ k^r=0 i^+3 j^+0 k^+λ i^+0 j^+2 k^+μ -2 i^-j^+k^We know that the equation r=a+λb+μc represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=0 i^+3 j^+0 k^; b=i^+0 j^+2 k^; c=-2 i^-j^+k^Normal vector, n=b×c=i^j^k^102-2-11=2 i^-5 j^-k^The vector equation of the plane in scalar product form isr. n=a. nr. 2 i^-5 j^-k^ = 0 i^+3 j^+0 k^. 2 i^-5 j^-k^r. 2 i^-5 j^-k^ = 0 - 15 + 0r. 2 i^-5 j^-k^ + 15 = 0

ii We know that the equation r=a+λb+μc represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a=2 i^+2 j^- k^; b=i^+2 j^+3 k^; c=5 i-2 j^+7 k^Normal vector, n=b×c=i^j^k^1235-27=20 i^+8 j^-12 k^The vector equation of the plane in scalar product form isr. n=a. nr. 20 i^+8 j^-12 k^=2 i^+2 j^- k^. 20 i^+8 j^-12 k^r. 4 5 i^+2 j^-3 k^=40+16+12r. 4 5 i^+2 j^-3 k^=68r. 5 i^+2 j^-3 k^=17



Page No 28.39:

Question 1:

Find the equation of the plane which is parallel to 2x − 3y + z = 0 and which passes through (1, −1, 2).

Answer:

Let the equation of a plane parallel to the given plane be 2x-3y+z=k ... 1This passes through (1, -1, 2). So, 2 1-3 -1+2=kk=7Substituting this in (1), we get2x-3y+z=7, which is the equation of the required plane.

Page No 28.39:

Question 2:

Find the equation of the plane through (3, 4, −1) which is parallel to the plane r·2i^-3j^+5k^+2=0.

Answer:

Let the equation of a plane parallel to the given plane be r. 2i^-3j^+5k^ = k ... 1xi^+yj^+zk^. 2i^-3j^+5k^ = k This passes through (3, 4, -1).  So, 3i^+4j^-k^. 2i^-3j^+5k^ = k k = 6-12-5 = -11Substituting this in (1), we getr. 2i^-3j^+5k^=-11r. 2i^-3j^+5k^+11=0,  which is the equation of the required plane.

Page No 28.39:

Question 3:

Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3).

Answer:

The equation of the plane passing through the line of intersection of the given planes is2x-7y+4z-3+λ 3x-5y+4z+11=0 ... 1This passes through (-2, 1, 3). So,-4-7+12-3+λ -6-5+12+11=0-2+12λ=0λ=16Substituting this in (1), we get2x-7y+4z-3+16 3x-5y+4z+11=012x-42y+24z-18+3x-5y+4z+11=015x-47y+28z=7

Page No 28.39:

Question 4:

Find the equation of the plane through the point 2i^+j^-k^ and passing through the line of intersection of the planes r·i^+3j^-k^=0 and r·j^+2k^=0.

Answer:

The equation of the plane passing through the line of intersection of the given planes isr. i^+3 j^-k^+λ r. j^+2 k^=0 r. i^+3+λ j^+-1+2λk^=0... 1This passes through 2 i^+j^-k^. So,2 i^+j^-k^ i^+3+λ j^+-1+2λk^=02+3+λ+1-2λ=0λ=6Substituting this in (1), we getr. i^+3+6 j^+-1+12k^=0r. i^+9 j^+11 k^=0

Page No 28.39:

Question 5:

Find the equation of the plane passing through the line of intersection of the planes 2xy = 0 and 3zy = 0 and perpendicular to the plane 4x + 5y − 3z = 8.

Answer:

The equation of the plane passing through the line of intersection of the given planes is2x-y+λ 3z-y=0 2x+-1-λy+3λ z=0... 1This plane is perpendicular to 4x+5y-3z=8. So,2 4+-1-λ 5-9λ=0 (Because a1a2+b1b2+c1c2=0)8-5-5λ-9λ=0λ=314Substituting this in (1), we get2x+-1-314y+3314 z=028x-17y+9z=0

Page No 28.39:

Question 6:

Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z − 4 = 0 and 2x + yz + 5 = 0 and which is perpendicular to the plane 5x + 3y − 6z + 8 = 0.

Answer:

The equation of the plane passing through the line of intersection of the given planes isx + 2y + 3z - 4 + λ 2x + y - z + 5 = 0 1 + 2λx + 2 + λy + 3 - λz - 4 + 5λ = 0... 1This plane is perpendicular to 5x + 3y - 6z + 8 = 0. So,5 1 + 2λ + 32 + λ - 6 3 - λ = 0 (Because a1a2+b1b2+c1c2=0)5 + 10λ + 6 + 3λ - 18 + 6λ = 019λ - 7 = 0λ = 719Substituting this in (1), we get1 + 2 719x +2 + 719y + 3 - 719z - 4 + 5 719 =  033x + 45y + 50z - 41=0

Page No 28.39:

Question 7:

Find the equation of the plane through the line of intersection of the planes x + 2y + 3z + 4 = 0 and xy + z + 3 = 0 and passing through the origin.

Answer:

The equation of the plane passing through the line of intersection of the given planes isx + 2y + 3z + 4 + λ x - y + z + 3 = 0 ... 1This passes through (0, 0, 0). So,0 + 0 + 0 + 4 + λ 0 - 0 + 0 + 3 = 04 + 3λ = 0λ = -4 3Substituting this in (1), we getx + 2y + 3z + 4 - 43x - y + z + 3 = 0 -x + 10y + 5z = 0x - 10y - 5z = 0

Page No 28.39:

Question 8:

Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x − 3y + 2z − 5 = 0 and 2xy + 3z − 1 = 0 and passing through (1, −2, 3).

Answer:

The equation of the plane passing through the line of intersection of the given planes isx - 3y + 2z - 5 + λ 2x - y + 3z - 1 = 0 ... 1This passes through (1, -2, 3).  So,1 + 6 + 6 - 5 + λ 2 + 2 + 9 - 18 + 12λ = 0λ = -2 3Substituting this in (1), we getx - 3y + 2z - 5 - 23 2x - y + 3z - 1 = 0-x - 7y - 13 = 0x + 7y + 13 = 0r. i^ + 7 j^ + 13 = 0, which is the required vector equation of the plane.

Page No 28.39:

Question 9:

Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z − 4 = 0, 2x + yz + 5 = 0.

Answer:

The equation of the plane passing through the line of intersection of the given planes isx + 2y + 3z - 4 + λ 2x + y - z + 5 = 0 1 + 2λx + 2 + λy + 3 - λz - 4 + 5λ = 0... 1This plane is perpendicular to 5x + 3y + 6z + 8 = 0. So,5 1 + 2λ + 3 2 + λ + 6 3 - λ = 0 (Because a1a2+b1b2+c1c2=0)5 + 10λ + 6 + 3λ + 18 - 6λ = 07λ + 29 = 0λ = -297Substituting this in (1), we get1 + 2 -297x + 2 - 297y + 3 + 297z - 4 + 5 -297 = 0-51x - 15y + 50z - 173  = 0 51x + 15y - 50z + 173 = 0

Page No 28.39:

Question 10:

Find the equation of the plane through the line of intersection of the planes r·i^+3j^+6=0 and r·3i^-j^-4k^=0, which is at a unit distance from the origin.

Answer:

The equation of the plane passing through the line of intersection of the given planes isr. i^ + 3 j^ + 6 + λ r. 3i^ - j^ - 4 k^ = 0 r. 1 + 3λ i^ + 3 - λ j^ - 4λ k^ + 6 = 0... 1r. 1 + 3λ i^ + 3 - λ j ^- 4λ k^ = -6r. -1 - 3λ i^ + λ - 3 j^ + 4λ k^ = 6Dividing both sides by -1-3λ2+λ-32+16λ2, we getr. -1 - 3λ i^ + λ - 3 j^ + 4λ k^-1 - 3λ2 + λ - 32 + 16λ2 = 6-1 - 3λ2 + λ - 32 + 16λ2, which is the normal form of plane (1), wherethe perpendicular distance of plane (1) from the origin = 6-1 - 3λ2 + λ - 32 + 16λ21=6-1 - 3λ2 + λ - 32 + 16λ2 (Given)-1 - 3λ2 + λ - 32 + 16λ2 = 61 + 9λ2 + 6λ + λ2 + 9 - 6λ + 16λ2 = 3626λ2 - 26 = 0λ2 = 1λ = 1 , -1Case 1: Substituting λ = 1 in (1), we getr. 4 i ^+ 2 j ^- 4 k^ + 6 = 0Case 2: Substituting λ=-1 in (1), we getr. -2 i^ + 4 j^ + 4 k^ + 6 = 0

Page No 28.39:

Question 11:

Find the equation of the plane passing through the intersection of the planes 2x + 3yz + 1 = 0 and x + y − 2z + 3 = 0 and perpendicular to the plane 3xy − 2z − 4 = 0.

Answer:

The equation of the plane passing through the line of intersection of the given planes is2x + 3y - z + 1 + λ x + y - 2z + 3 = 0 2 + λx + 3 + λy + -1 - 2λz + 1 + 3λ = 0... 1This plane is perpendicular to 3x - y - 2z - 4 = 0. So,3 2 + λ - 3 + λ - 2 -1 - 2λ = 0 (Because a1a2 + b1b2 + c1c2 = 0)6 + 3λ - 3 - λ + 2 + 4λ  = 06λ + 5 = 0λ = -5   6 Substituting this in (1), we get2 - 56x + 3 - 56y + -1 - 2 -56z + 1 + 3 -56 = 07X + 13Y + 4z - 9 = 0

Page No 28.39:

Question 12:

Find the equation of the plane that contains the line of intersection of the planes r·i^+2j^+3k^-4=0 and r·2i^+j^-k^+5=0 and which is perpendicular to the plane r·5i^+3j^-6k^+8=0.

Answer:

The equation of the plane passing through the line of intersection of the given planes isr. i^+2 j^+3k^-4+λ r. 2i^+j^- k^+5=0 r. 1+2λ i^+2+λ j^+3-λk^-4+5λ=0... 1This plane is perpendicular to r. 5 i^+3 j^-6 k^+8=0. So,5 1+2λ+3 2+λ-6 3-λ=0 (Because a1a2+b1b2+c1c2=0)5+10λ+6+3λ-18+6λ=019λ-7=0λ=719Substituting this in (1), we getr. 1+2 719 i^+2+719 j^+3-719k^-4+5 719=0r. 33 i^+45 j^+50 k^-41=0x i^+y j^+z k^. 33 i^+45 j^+50 k^-41=033x+45y+50z-41=0

Page No 28.39:

Question 13:

Find the equation of the plane passing through (a, b, c) and parallel to the plane r·i^+j^+k^=2.

Answer:

Let the equation of a plane parallel to the given plane be r. i^ + j ^+ k^ = k xi^ + yj^ + zk^. i^ + j ^+ k^ = k ... 1This passes through (a, b, c). So, a i^ + b j^ + c k^. i^ + j^ + k^ = kk = a + b + cSubstituting this in (1), we getxi^ + yj^ + zk^. i ^+ j ^+ k^ = a + b + cx + y + z = a + b + c,  which is the equation of the required plane.

Page No 28.39:

Question 14:

Find the equation of the plane passing through the intersection of the planes r·2i^+j^+3k^=7, r·2i^+5j^+3k^=9 and the point (2, 1, 3).

Answer:

The equation of the plane passing through the line of intersection of the given planes isr. 2 i^ + j ^+ 3 k^ - 7 + λ r. 2 i^ + 5 j^ + 3 k^ - 9 = 0 r. 2 + 2λ i^ + 1 + 5λ j^ + 3 + 3λk^ - 7 - 9λ = 0... 1This passes through 2 i^ + j ^+ 3 k^. So,2 i^ + j^ + 3 k^ 2 + 2λ i^ + 1 + 5λ j ^+ 3 + 3λk^ - 7 - 9λ = 04 + 4λ + 1 + 5λ + 9 + 9λ - 7 - 9λ = 09λ + 7 = 0λ = -79Substituting this in (1), we getr. 2 + 2 -7 9 i^ + 1 + 5 -7 9 j^ + 3 + 3 -7 9k^ - 7 - 9 -7 9 = 0r. 4 i^ - 26 j^ + 6 k^ = 0r. 2 i^ - 13 j^ + 3 k^ = 0



Page No 28.4:

Question 1:

Find the equation of the plane passing through the following points.
(i) (2, 1, 0), (3, −2, −2) and (3, 1, 7)
(ii) (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6)
(iii) (1, 1, 1), (1, −1, 2) and (−2, −2, 2)
(iv) (2, 3, 4), (−3, 5, 1) and (4, −1, 2)
(v) (0, −1, 0), (3, 3, 0) and (1, 1, 1)

Answer:

(i) The equation of the plane passing through points (2, 1, 0), (3, −2, −2) and (3, 1, 7) is given by

x-2y-1z-03-2-2-1-2-03-21-17-0=0x-2y-1z-01-3-2107=0-21 x-2 - 9 y-1 + 3z = 0-21x + 42 - 9y + 9 + 3z = 0-21x - 9y + 3z + 51 = 021x + 9y - 3z = 517x + 3y - z = 17

(ii) The equation of the plane passing through points (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6) is given by

x+5y-0 z+6-3+5 10-0-9+6-2+56-0-6+6 = 0x+5yz+6210-3360 = 018 x+5 - 9y - 18 z+6 = 02 x+5 - y - 2 z+6 = 02x + 10 - y - 2z - 12 = 02x - y - 2z - 2 = 0

(iii) The equation of the plane passing through points  (1, 1, 1), (1, −1, 2) and (−2, −2, 2) is given by

x-1y-1z-11-1-1-12-1-2-1-2-12-1=0x-1y-1z-10-21-3-31=01 x-1-3 y-1-6 z-1 = 0x - 1 - 3y + 3 - 6z + 6  = 0x - 3y - 6z + 8 = 0

(iv) The equation of the plane passing through points (2, 3, 4), (−3, 5, 1) and (4, −1, 2) is given by

x-2y-3z-4-3-25-31-44-2-1-32-4=0x-2y-3z-4-52-32-4-2=0-16 x - 2 - 16 y - 3 + 16 z - 4 = 0x - 2 + y - 3 -  z - 4 = 0x + y - z = 1

(v) The equation of the plane passing through points (0, −1, 0), (3, 3, 0) and (1, 1, 1) is given by

x-0y+1z-03-03+10-01-01+11-0=0x-0y+1z-0340121=04x - 3 y + 1 + 2z = 04x - 3y + 2z = 3



Page No 28.40:

Question 15:

Find the equation of the plane through the intersection of the planes 3xy + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).

Answer:

The equation of the plane passing through the line of intersection of the given planes is3x - y + 2z - 4 + λ x + y + z - 2 = 0 ... 1This passes through (2, 2, 1). So,6 - 2 + 2 - 4 + λ 2 + 2 + 1 - 2 = 02 + 3λ = 0λ = -2 3Substituting this in (1), we get3x - y + 2z - 4 - 23 x + y + z - 2 = 09x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 07x - 5y + 4z = 8

Page No 28.40:

Question 16:

Find the vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane xy + z = 0.

Answer:

The equation of the plane passing through the line of intersection of the given planes isx + y + z - 1 + λ 2x + 3y + 4z - 5 = 0 1 + 2λx + 1 + 3λy + 1 + 4λz - 1 - 5λ = 0... 1This plane is perpendicular to x - y + z = 0. So,1 + 2λ - 1 1 + 3λ + 1 + 4λ = 0 (Because a1a2 + b1b2 + c1c2 = 0)1 + 2λ - 1 - 3λ + 1 + 4λ = 03λ + 1 = 0λ = -1 3Substituting this in (1), we get1 + 2 -1 3x + 1 + 3 -1  3y + 1 + 4 -1  3z - 1 - 5 -1  3 = 0x - z + 2 = 0

Page No 28.40:

Question 17:

Find the equation of the plane passing through the intersection the planers
r.i^ + j^ + k^ = 6 and r. 2i^ + 3j^ +4k^ = -5 and the point (1, 1, 1).

Answer:

The equation of the plan passing through the line of intersection of the given planes is 

r.i^+j^+k^ - 6 + λr.2 i^+3 j^+4 k^+5=0r.1 + 2λ i^ + 1 + 3λ j^ + 1 + 4λ k^- 6 + 5λ = 0...1

This passes through i^ + j^ + k^. So,
i^ + j ^ + k^. 1 + 2λ i^ + 1 + 3λ j^ + 1 + 4λ k^ - 6 + 5λ = 01 + 2λ + 1 + 3λ + 1 + 4λ - 6 + 5λ = 014λ - 3 = 0λ = 314

Substituting this in (1), we get
r.1 + 2314 i^ + 1 + 3314 j^ + 1 + 4314 k^ - 6 + 5 314 = 0r.20 i^ + 23j^ + 26k^ = 69

Page No 28.40:

Question 18:

Find the equation of the plane which contains the line of intersection of the planes x+2y+3z -4=0 and 2x+y-z + 5=0 and whose x-intercept is twice its z-intercept. Hence, write the equation of the plane passing through the point (2, 3, -1) and parallel to the plane obtained above.

Answer:


The equation of the family of planes passing through the intersection of the planes x + 2y + 3z − 4 = 0 and 2xy − z + 5 = 0 is

(x + 2y + 3z − 4) + k(2x + y − z + 5) = 0, where k is some constant

2k+1x+k+2y+3-kz=4-5kx4-5k2k+1+y4-5kk+2+z4-5k3-k=1
It is given that x-intercept of the required plane is twice its z-intercept.

4-5k2k+1=24-5k3-k4-5k3-k=4k+24-5k4-5k3-k-4k-2=04-5k1-5k=0
4-5k=0 or 1-5k=0k=45 or k=15
When k=45, the equation of the plane is 2×45+1x+45+2y+3-45z=4-5×4513x+14y+11z=0.

This plane does not satisfies the given condition, so this is rejected.

When k=15, the equation of the plane is 2×15+1x+15+2y+3-15z=4-5×157x+11y+14z=15.

Thus, the equation of the required plane is 7x + 11y + 14z = 15.

Also, the equation of the plane passing through the point (2, 3, −1) and parallel to the plane 7x + 11y + 14z = 15 is

7x-2+11y-3+14z+1=07x+11y+14z=33

Page No 28.40:

Question 19:

Find the equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 and twice of its y-intercept is equal to three times its z-intercept. 

Answer:


The equation of the family of the planes passing through the intersection of the planes xyz = 1 and 2x + 3y + 4z = 5 is

(x + y + z − 1) + k(2x + 3y + 4z − 5) = 0, where k is some constant

⇒ (2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1                .....(1)

x5k+12k+ 1+y5k+13k+1+z5k+14k+1=1

It is given that twice of y-intercept is equal to three times its z-intercept.

25k+13k+1=35k+14k+15k+18k+2-9k-3=05k+1-k-1=05k+1k+1=0
5k+1=0 or k+1=0k=-15 or k=-1
Putting k=-15 in (1), we get

-25+1x+-35+1y+-45+1z=5×-15+13x+2y+z=0
This plane passes through the origin. So, the intercepts made by the plane with the coordinate axes is 0. Hence, this equation of plane is not accepted as twice of y-intercept is not equal to three times its z-intercept.

Putting k=-1 in (1), we get

-2+1x+-3+1y+-4+1z=5×-1+1-x-2y-3z=-4x+2y+3z=4
Here, twice of y-intercept is equal to three times its z-intercept.

Thus, the equation of the required plane is x + 2y + 3z = 4.
 



Page No 28.49:

Question 1:

Find the distance of the point 2i^-j^-4k^ from the plane r·3i^-4j^+12k^-9=0.

Answer:

We know that the perpendicular distance of a point P of position vector a from the plane rn = d is given by p = a. n-dnHere,a = 2 i^ - j^ - 4 k^; n = 3 i^ - 4 j^ + 12 k^; d = 9So, the required distance, p=2 i^ - j ^- 4 k^. 3 i^ - 4 j^ + 12 k^ - 93 i^ - 4 j^ + 12 k^=6 + 4 - 48 - 99 + 16 + 144=-4713=4713 units

Page No 28.49:

Question 2:

Show that the points i^-j^+3k^ and 3i^+3j^+3k^ are equidistant from the plane r·5i^+2j^-7k^+9=0.

Answer:

The given plane isr. 5 i^ + 2 j ^- 7 k^ + 9 = 0r. -5 i ^- 2 j^ + 7 k^ = 9We know that the perpendicular distance of a point P of position vector a from the plane rn  =  d is given by p = a. n-dnFinding the distance from i^-j^+3 k^ to the given planeHere,a = i^ - j^ + 3 k^; n = -5 i^ - 2 j ^+ 7 k^; d = 9So, the required distance p is given byp= i^ - j^ + 3 k^. -5 i ^- 2 j ^+ 7 k^ - 9-5 i^ - 2 j^ + 7 k^= -5 + 2 + 21 - 925 + 4 + 49=    9 78= 978 units ... (1)Finding the distance from 3 i^+3 j^+3 k^ to the given planeHere, a = 3 i^ + 3 j^ + 3 k^; n = -5 i ^- 2 j^ + 7 k^; d = 9So, the required distance p is given byp=3 i^ + 3 j ^+ 3 k^. -5 i^ - 2 j^ + 7 k^ - 93 i ^+ 3 j^ + 3 k^=-15 - 6 + 21 - 925 + 4 + 49=-978=978 units ... (2)From (1) and (2), we can say that the given points are equidistant from the given plane.

Page No 28.49:

Question 3:

Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0.

Answer:

We know that the distance of the point x1, y1, z1 from the plane ax + by + cz + d = 0 is given byax1 + by1 + cz1 + da2 + b2 + c2So, the required distance = 2 + 2 3 - 2 -5 - 912 + 22 + -22= 2 + 6 + 10 - 91 + 4 + 4= 93= 3  units

Page No 28.49:

Question 4:

Find the equations of the planes parallel to the plane x + 2y − 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).

Answer:

The equation of the plane parallel to the given plane isx + 2y - 2z + k = 0 ... 1It is given that plane (1) is at a distance of 2 units from (2, 1, 1).2 + 2 - 2 + k12 + 22 + -22 = 2 2 + k 3 = 22 + k = 62 + k = 6; 2 + k = -6k = 4; k = -8Substituting these two values one by one in (1), we getx + 2y - 2z + 4 = 0 andx + 2y - 2z - 8 = 0, which are the equations of the required planes.

Page No 28.49:

Question 5:

Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0.

Answer:

We know that the distance of the point x1, y1, z1 from the plane ax+by+cz+d=0 is given byax1+by1+cz1+da2+b2+c2Distance of the point (1, 1, 1) from the plane 3x + 4y - 12z + 13 = 0The required distance =3 1 + 4 1 - 12 1 + 1332 + 42 + -122=3 + 4 - 12 + 139 + 16 + 144=813 units ... (1)Distance of the point (-3, 0, 1) from the plane 3x + 4y - 12z + 13 = 0The required distance =3 -3 + 4 0 - 12 1 + 1332 + 42 + -122=-9+0-12+139+16+144=813 units .... (2)From (1) and (2), we can say that the given points are equidistant from the given plane.

Page No 28.49:

Question 6:

Find the equations of the planes parallel to the plane x − 2y + 2z − 3 = 0 and which are at a unit distance from the point (1, 1, 1).

Answer:

The equation of the plane parallel to the given plane isx - 2y + 2z + k = 0 ... 1It is given that plane (1) is at a distance of 1 unit from (1, 1, 1).1 - 2 + 2 + k12 + -22 + 22 = 11 + k3 = 11 + k = 31 + k = 3; 1 + k = -3k = 2; k = -4Substituting these two values one by one in (1), we getx - 2y + 2z + 2 = 0 andx - 2y + 2z - 4 = 0, which are the equations of the required planes.

Page No 28.49:

Question 7:

Find the distance of the point (2, 3, 5) from the xy - plane.

Answer:

We know that the distance of the point x1, y1, z1 from the plane ax + by + cz + d = 0 is given byax1 + by1 + cz1 + da2 + b2 + c2Equation of the xy- plane is z = 0, which means 0x + 0y + z = 0So, the required distance = 0 2+0 3+ 502+02+12=  51= 5 units

Page No 28.49:

Question 8:

Find the distance of the point (3, 3, 3) from the plane r·5i^+2j^-7k+9=0

Answer:

The given plane isr. 5 i^ + 2 j^ - 7 k^ + 9 = 0r. -5 i^ - 2 j^ + 7 k^  =  9We know that the perpendicular distance of a point P of position vector a from the plane rn = d is given by p = a. n - dnFinding the distance from (3, 3, 3) (which means 3 i^+3 j^+3 k^) to the given planeHere, a = 3 i^ + 3 j^ + 3 k^; n = -5 i^ - 2 j^ + 7 k^; d = 9So, the required distance p = 3 i^ + 3 j^ + 3 k^. -5 i^ - 2 j^ + 7 k^ - 93 i^ + 3 j^ + 3 k^= -15 - 6 + 21 - 925 + 4 + 49=  -9 78=   9 78 units

Page No 28.49:

Question 9:

If the product of the distances of the point (1, 1, 1) from the origin and the plane xy + z + λ = 0 be 5, find the value of λ.

Answer:

We know that the distance of the point x1, y1, z1 from the plane ax+by+cz+d=0 is given byax1+by1+cz1+da2+b2+c2Distance of the point (1, 1, 1) from the plane x-y+z+λ=0The required distance =1-1+1+λ12+-12+12=1+λ3 units ... (1)Distance of the point (0, 0, 0) from the plane x-y+z+λ=0The required distance =0-0+0+λ12+-12+12=λ3 units ... (2)It is given that the product of the distances (1) and (2) is 5.1+λ3 ×λ3 =5λ2+λ-15=0

Disclaimer: The answer or problem given in the text book is incorrect for this.

Page No 28.49:

Question 10:

Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7.

Answer:

Let x, y be a point which is equidistant from the given planes. Then,3x - 4y + 12z - 69 + 16 + 144 = 4x + 3z - 716 + 93x - 4y + 12z - 613 = ±4x + 3z - 7515x - 20y + 60z - 30  =  52x + 39z - 91; 15x - 20y + 60z - 30  =  -52x - 39z + 9137x + 20y - 21z - 61 = 0; 67x - 20y + 99z  =  121

Page No 28.49:

Question 11:

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).         [CBSE 2014]

Answer:


The given points are A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).

The equation of the plane ABC is given by

x-x1y-y1z-z1x2-x1y2-y1z2-z1x3-x1y3-y1z3-z1=0x-2y-5z--3-2-2-3-55--35-23-5-3--3=0x-2y-5z+3-4-883-20=0x-2y-5z+312-23-20=0-4x-2-6y-5-8z+3=02x-2+3y-5+4z+3=02x+3y+4z-7=0

∴ Distance between the point (7, 2, 4) and the plane 2x+3y+4z-7=0

= Length of perpendicular from (7, 2, 4) to the plane 2x+3y+4z-7=0

=2×7+3×2+4×4-722+32+42=14+6+16-74+9+16=2929=29 units

Thus, the required distance between the given point and the plane is 29 units.

Page No 28.49:

Question 12:

A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.       [CBSE 2014]

Answer:


We know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is xa+yb+zc=1.

So, the equation of the plane which makes intercepts −6, 3, 4 on the x-axis, y-axis and z-axis, respectively is

x-6+y3+z4=1-2x+4y+3z=122x-4y-3z+12=0

∴ Length of the perpendicular from (0, 0, 0) to the plane 2x-4y-3z+12=0

 =2×0-4×0-3×0+1222+-42+-32=124+16+9=1229 units

Thus, the length of the perpendicular from the origin to the plane is 1229 units.

Page No 28.49:

Question 13:

Find the distance of the point (1, -2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes x-y+2z=3 and 2x-2y+z+12=0.

Answer:


Let the equation of plane passing through the point (1, 2, 2) be

ax-1+by-2+cz-2=0          .....(1)

Here, abc are the direction ratios of the normal to the plane.

The equations of the given planes are x − y + 2z = 3 and 2x − 2y + z + 12 = 0.

Plane (1) is perpendicular to the given planes.

a − b + 2c = 0           .....(2)

2a − 2b + c = 0         .....(3)

Eliminating ab and c from (1), (2) and (3), we get

x-1y-2z-21-122-21=0x-1-1+4-y-21-4+z-2-2+2=03x+3y-9=0x+y-3=0
∴ Distance of the point (1, −2, 4) from the plane xy − 3 = 0

=1-2-312+12+02=-42=22 units



Page No 28.5:

Question 2:

Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane.

Answer:

The equation of the plane passing through the points (0, −1, −1), (4, 5, 1) and (3, 9, 4) is given by

x-0y+1z+14-05+11+13-09+14+1=0xy+1z+14623105=010x - 14 y + 1 + 22 z + 1 = 05x - 7 y + 1 + 11 z + 1 = 05x - 7y + 11z + 4 = 0Substituting the last point (-4, 4, 4) (it means x = -4;  y = 4;  z = 4) in this plane equation, we get 5 -4 - 7 4 + 11 4 + 4 = 0- 48 + 48 = 00 = 0So, the plane equation is satisfied by the point (-4, 4, 4).So, the given points are coplanar and the equation of the common plane (as we already found) is5x - 7y + 11z + 4 = 0.

Page No 28.5:

Question 3:

Show that the following points are coplanar.
(i) (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0)
(ii) (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)

Answer:

(i) The equation of the plane passing through points (0, −1, 0), (2, 1, −1), (1, 1, 1) is given by

x-0y+1z-02-01+1-1-01-01+11-0=0xy+1z22-1121=04x-3 y+1+2 z=04x-3y+2z-3=0Substituting the last point (3, 3, 0) (it means x = 3; y = 3; z = 0) in this plane equation, we get 4 3-3 3+2 0-3 = 012-12 = 00 = 0So, the plane equation is satisfied by the point (3, 3, 0).So, the given points are coplanar.

(ii) The equation of the plane passing through (0, 4, 3), (−1, −5, −3), (−2, −2, 1) is
x-0y-4z-3-1-0-5-4-3-3-2-0-2-41-3=0xy-4z-3-1-9-6-2-6-2=0-18x+10 y-4-12 z-3=09x-5 y-4+6 z-3=09x-5y+6z+2=0Substituting the last point (1, 1, -1) (it means x = 1; y = 1; z=-1) in this plane equation, we get 9 1-5 1+6 -1+2=04-4=00=0So, the plane equation is satisfied by the point (1, 1, -1).So, the given points are coplanar.

Page No 28.5:

Question 4:

Find the coordinates of the point where the line through (3,-4,-5) and  B (2,-3,1) crosses the plane passing through three points L(2,2,1), M(3,0,1) and N(4,-1,0). Also, find the ratio in which diveides the line segment AB.
 

Answer:


Equation of the plane passing through the points L(2, 2, 1),  M(3, 0, 1) and N(4, −1, 0) is

r-2i^+2j^+k^.i^-2j^×i^-j^-k^=0r-2i^+2j^+k^.2i^+j^+k^=0r.2i^+j^+k^=2i^+2j^+k^.2i^+j^+k^r.2i^+j^+k^=4+2+1=7         .....1
The equation of line segment through A(3, −4, −5) and B(2−3, 1) is

x-32-3=y+4-3+4=z+51+5i.e. x-3-1=y+41=z+56
Any point on this line is of the form -λ+3, λ-4, 6λ-5.

This point lies on the plane (1).

-λ+3i^+λ-4j^+6λ-5k^.2i^+j^+k^=72-λ+3+λ-4+6λ-5=75λ=10λ=2
Thus, the coordinates of the point P are (−2 + 3, 2 − 4, 6 × 2 − 5) i.e. (1, −2, 7).

Suppose P divides the line segment AB in the ratio μ : 1.

1,-2,7=2μ+3μ+1,-3μ-4μ+1,μ-5μ+12μ+3μ+1=1, -3μ-4μ+1=-2, μ-5μ+1=72μ+3=μ+1, -3μ-4=-2μ-2, μ-5=7μ+7μ=-2
Thus, the point P divides the line segment AB externally in the ratio 2 : 1.



Page No 28.51:

Question 1:

Find the distance between the parallel planes 2xy + 3z − 4 = 0 and 6x − 3y + 9z + 13 = 0.

Answer:

Multiplying the first equation of the plane by 3, we get6x - 3y + 9z - 12 = 0 6x - 3y + 9z = 12 ... 1The second equation of the plane is6x-3y+9z=-13 ... 2We know that the distance between two planes ax+by+cz=d1 and ax+by+cz=d2 is d2-d1a2+b2+c2So, the required distance -13-1262+-32+92=-2536+9+81=  25126=53 14 units

Page No 28.51:

Question 2:

Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes.

Answer:

Let the equation of a plane parallel to the given plane be 2x - 3y + 5z = k ... 1This passes through (3, 4, -1).  So, 2 3 -3 4+5 -1 = kk = -11Substituting this in (1), we get2x - 3y + 5z =-11... 1, which is the equation of the required plane.The equation of the given plane is2x - 3y + 5z = -7 ... 2We know that the distance between two planes ax+by+cz=d1 and ax+by+cz=d2 is d2-d1a2+b2+c2So, the required distance =-7 - -1122 + -32 + 52=-7 + 114 + 9 + 25= 438 units

Page No 28.51:

Question 3:

Find the equation of the plane mid-parallel to the planes 2x − 2y + z + 3 = 0 and 2x − 2y + z + 9 = 0.

Answer:

We know that the distance between two planes ax+by+cz=d1 and ax+by+cz=d2 is d2-d1a2+b2+c2The equation of plane that is mid-parallel to the planes 2x-2y+z+3=0 ... 12x-2y+z+9=0 ... 2is of the form 2x-2y+z+k=0 ... 3It means that the distance between (1) and (3) = distance between (1) and (2) k - 3  4 + 4 + 1 =  k - 9 4 + 4 + 1k-3 = k-9k-3 = k-9  or k-3 = - k-93 = 9 (false); k-3  = -k + 92k = 12k = 6Substituting this in (3), we get 2x-2y+z+6=0, which is the required equation of the plane.

Page No 28.51:

Question 4:

Find the distance between the planes r·i^+2j^+3k^+7=0 and r·2i^+4j^+6k^+7=0.

Answer:

The given planes arer. i^+2 j^+3 k^ = -7 x+2y+3z = -7 Multiplying this equation of the plane by 2, we get2x+4y+6z=-14 ... (1)andr. 2 i^ + 4 j^ + 6 k^ = -72x + 4y + 6z = -7 ... 2We know that the distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is  d2-d1a2+b2+c2So, the required distance -7 - -1422 + 42 + 62= 7 4 + 16 + 36=   756 units



Page No 28.61:

Question 1:

Find the angle between the line r=2i^+3j^+9k^+λ2i^+3j^+4k^ and the plane r·i^+j^+k^=5.

Answer:

We know that the angle θ between the line r = a + λ b and the plane rn=d is given bysin θ = b. nb n.Here, b = 2 i^+3 j^+4 k^ and n = i^+j^+k^So, sin θ = 2 i^+3 j^+4 k^. i^+j^+k^2 i^+3 j^+4 k^ i^+j^+k^ = 2+3+44+9+16 1+1+1 = 929 3 = 3 329θ = sin-1 3 329

Page No 28.61:

Question 2:

Find the angle between the line x-11=y-2-1=z+11 and the plane 2x + yz = 4.

Answer:

The given line is parallel to the vector b = i^ - j ^+ k^ and the given plane is normal to the vector n=2 i^ + j ^- k^.We know that the angle θ between the line and the plane  is given bysin θ = b. nb n= i^-j^+k^. 2 i^+j^-k^i^-j^+k^ 2 i^+j^-k^ = 2-1-11+1+1 4+1+1 = 0θ=sin-1 0=0

Page No 28.61:

Question 3:

Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3xy + z = 1.

Answer:

It is given that the line passes through A 3, -4, -2 and B 12, 2, 0.So, b = AB=OB-OA = 12 i^+2 j^+0 k^-3 i^-4 j^-2 k^ = 9 i^+6 j^+2 k^The given line is parallel to the vector b=9 i^+6 j^+2 k^ and the given plane is normal to the vector n=3 i^-j^+k^.We know that the angle θ between the line and the plane  is given bysin θ=b. nb n=9 i^+6 j^+2 k^. 3 i^-j^+k^9 i^+6 j^+2 k^ 3 i^-j^+k^ = 27 - 6 + 281 + 36 + 4 9 + 1 + 1 = 2311 11θ=sin-1 2311 11

Page No 28.61:

Question 4:

The line r=i^+λ2i^-mj^-3k^ is parallel to the plane r·mi^+3j^+k^=4. Find m.

Answer:

The given line is parallel to the vector b = 2 i^ - m j^ - 3 k^ and the given plane is normal to the vector n = m i^ + 3 j^ + k^.If the line is parallel to the plane, the normal to the plane is perpendicular to the line.b  nb. n = 02 i^ - m j^ - 3 k^. m i^ + 3 j^ + k^ = 02m - 3m - 3 = 0-m - 3 = 0 m = -3

Page No 28.61:

Question 5:

Show that the line whose vector equation is r=2i^+5j^+7k^+λi^+3j^+4k^ is parallel to the plane whose vector equation is r·i^+j^-k^=7. Also, find the distance between them.

Answer:

The given plane passes through the point with position vector a=2 i^+5 j^+7 k^ and is parallel to the vector b=i^+3 j^+4 k^. The given plane is ri ^+ j^-k^ = 7 or . ccSo, the normal vector, n = i^+j^- k^ and d=7.Now, b. n=i^ + 3 j^ + 4 k^. i ^+ j ^- k^ = 1 + 3 - 4 = 4 - 4 = 0So, b is perpendicular to n.So, the given line is parallel to the given plane.The distance between the line and the parallel plane . Then,d = length of the perpendicular from the point  a=2 i^+5 j^+7 k^ to the plane rn = dd=a. n-dn=2 i^ + 5 j ^+ 7 k^. i^ + j ^- k^ - 7  i^ + j ^- k ^=2+5-7-71+1+1= 73 units

Page No 28.61:

Question 6:

Find the vector equation of the line through the origin which is perpendicular to the plane r·i^+2j^+3k^=3.

Answer:

The required line is perpendicular to the plane r. i^ - 2 j ^+ 3 k^ = 3. Therefore, it is parallel to the normal i^ + 2 j^ + 3 k^. Thus, the required line passes through the point with position vector a = 0 i^+0 j^+0 k^ and is parallel to the vector n = i^ - 2 j^ + 3 k^.So, its vector equation isr = a + λ n^r = 0 i^ + 0 j^ + 0 k ^+ λ i^ - 2 j^ + 3 k^r= λ i^ - 2 j ^+ 3 k^

Page No 28.61:

Question 7:

Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.

Answer:

The equation of the plane through (2, 3, -4) isa x-2 + b y-3 + c z+4 = 0 ... 1This plane passes through (1, -1, 3).  So,a 1 - 2 + b -1 - 3 + c 3 + 4=0-a - 4b + 7c = 0  ... 2Again plane (1) is parallel to x-axis.  It means that plane (1) is perpendicular to the yz-plane whose equation is x = 0 or 1. x + 0. y + 0. z = 0a 1 + b 0 + c 0 = 0  ... 3 (Because a1a2+b1b2+c1c2=0)Solving (1), (2) and (3), we getx-3y-3z+4-1-47100=00 x - 3 + 7 y - 3 + 4 z + 4 = 07y + 4z - 5 = 0

Page No 28.61:

Question 8:

Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line x-41=y+3-4=z+17.

Answer:

The equation of the plane through (0,0, 0) isa x-0+b y-0+c z-0=0 ax+by+cz=0... 1This plane passes through (3, -1, 2). So,3a-b+2c=0  ... 2Again plane (1) is parallel to the given line. It means that the normal to plane (1) is perpendicular to the line.a 1+b -4+c 7=0  ... 3 (Because a1a2+b1b2+c1c2=0)Solving (1), (2) and (3), we getx y z3-1 21-4 7 = 0x - 19y - 11z = 0

Page No 28.61:

Question 9:

Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes r·i^-j^+2k^=5 and r·3i^+j^+2k^=6

Answer:

Let the direction ratios of the required line be proportional to a, b, c. As it passes through (1, 2, 3), its equations arex-1a = y-2b = z-3c ... 1It is given that (1) is parallel to the planes ri^-j^+2 k^ = 5 and r3 i^+j^+2 k^ = 6 or x - y + 2z = 5 and 3x + y + 2z = 6a - b + 2c = 0 ... 23a + b + 2z = 0 ... 3Solving these two by cross-multiplication method, we geta-2-2 = b6-2 = c1+3a-4 = b4 = c4a1 = b-1 = c-1 = λ(say)a = λ; b = -λ; c = -λSubstituting these values in (1), we getx-11=y-2-1=z-3-1, which is the Cartesian form of the required line.Vector formThe given line passes through a point whose position vector is a=i^+2 j^+3 k^ and is parallel to the vector b = i^- j ^+ k^. So, its equation in vector form isr=a+λbr=i^+2 j^+3 k^+λi^-j^+k^

Disclaimer: The answer given for this problem in the text book is incorrect. The problem should be same as problem #19 to get the text book answer.

Page No 28.61:

Question 10:

Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0.

Answer:

Let a, b, c be the direction ratios of the line of section of the given planes.As this line lies on both the planes, their normals are perpendicular to it.5a+2b-4c=0 ...12a+8b+2c=0a+4b+c=0 ...2Using cross-multiplication method, we geta2+16=b-4-5=c20-2a18=b-9=c18a2=b-1=c2So, the direction ratios of the line are proportional to 2, -1, 2.Direction ratios of the given line are 4,-2, -5.Now,2 4+-1 -2+2 -5=8+2-10=0So, the line of section of the given planes is parallel to the given plane. 

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Question 11:

Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2xy + 3z − 5 = 0.

Answer:

Let a, b, c be the direction ratios of the given line.Since the line passes through the point (1, -1, 2) isx-1a = y+1b = z-2c ...1Since this line is perpendicular to the plane 2x-y+3z-5=0, the line is parallel to the normal of the plane.So, the direction ratios of the line are proportional to the direction ratios of the given plane.So, a2 = b-1 = c3 = λa = 2λ;  b = -λ;  c = 3λSubstituting these values in (1), we getx-12=y+1-1=z-23, which is the Cartesian form of the line.Vector formThe given line passes through a point whose position vector is a = i ^-  j^ + 2 k^ and is parallel to the vector b  =  2 i ^- j ^+ 3 k^So, its equation in vector form isr = a + λbr = i^ -  j^ + 2 k^ + λi^ j^ + 3 k^

Page No 28.61:

Question 12:

Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

Answer:

The equation of the plane through (2, 2, -1) isa x - 2 + b y - 2 + c z + 1 = 0 ... 1This plane passes through (3, 4, 2).  So,a 3 - 2 + b 4 - 2 + c 2 + 1 = 0a + 2b + 3c = 0 ... 2 Again plane (1) is parallel to the line whose direction ratios are 7, 0, 6.It means that the normal of plane (1) is perpendicular to the line whose direction ratios are 7, 0, 6.7a+0b+6c=0 (Because a1a2+b1b2+c1c2=0)Solving (1), (2) and (3), we getx-2y-2z+1123706=012 x-2+15 y-2-14 z+1=012x+15y-14z-68=0

Page No 28.61:

Question 13:

Find the angle between the line x-23=y+1-1=z-32 and the plane 3x + 4y + z + 5 = 0.

Answer:

The given line is parallel to the vector b=3 i^-j^+2 k^ and the given plane is normal to the vector n=3 i^+4 j^+k^.We know that the angle θ between the line and the plane is given bysin θ=b. nb n=3 i^-j^+2 k^. 3 i^+4 j^+k^3 i^-j^+2 k^ 3 i^+4 j^+k^ = 9-4+29+1+4 9+16+1 = 714 26 =72 7 2 13 = 752=752θ=sin-1 752

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Question 14:

Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane.

Answer:

The equation of the plane passing through the intersection of the given planes isx-2y+z-1+λ 2x+y+z-8 = 01+2λ x+-2+λ y+1+λ z-1-8λ = 0 ... 1This plane is parallel to the line whose direction ratios are proportional to 1,2,1.So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to 1, 2, 1.1+2λ 1+-2+λ 2+1+λ 1 = 01+2λ-4+2λ+1+λ = 05λ-2 = 0λ = 25Substituting this in (1), we get1+2 25 x+-2+25 y+1+25 z-1-8 25 =09x-8y+7z-21=0... 2, which is the required equation of the plane.Perpendicular distance of plane (2) from (1, 1, 1)=9 1 - 8 1 + 7 1 - 2192 + -82 + 72=  -13194=13194 units

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Question 15:

State when the line r=a+λb is parallel to the plane r·n=d. Show that the line r=i^+j^+λ3i^-j^+2k^ is parallel to the plane r·2j^+k^=3. Also, find the distance between the line and the plane.

Answer:

The given plane passes through the point with position vector a= i^+j^+0 k^ and is parallel to the vector b=3 i^- j^+2 k^. The given plane is r2j^+k^=3 or rn=dSo, normal vector, n=0 i^+2 j^+k^ and d=3Now, b. n=3 i^- j^+2 k^. i^+2 j^+k^=0-2+2=0So, b is perpendicular to n.Hence,the given line is parallel to the given plane.The distance between the line and the parallel plane is the distance between any point on the line and the given plane. The plane passes through the point a= i^+j^+0 k^.The perpendicular distance from the given line to the plane isd=a. n-dn=i^ + j^ + 0 k^. i^ + 2 j^ + k^ -3i^ + 2 j^ + k^=0 + 2 + 0 -30 + 22 + 12=  15 units

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Question 16:

Show that the plane whose vector equation is r·i^+2j^-k^=1 and the line whose vector equation is r=-i^+j^+k^+λ2i^+j^+4k^ are parallel. Also, find the distance between them.

Answer:

The given plane passes through the point with position vector a=- i^ + j^ + k^ and is parallel to the vector b = 2 i^+j^+4 k^. The given plane is ri^ + 2 j^ - k^ = 1 or  rn = d.So, normal vector, n = i^+2 j^-k^ and d = 1Now, b. n = 2 i^ + j^ + 4 k^. i^ + 2 j ^- k^ = 2 + 2 - 4 = 0So, b is perpendicular to n.So, the given line is parallel to the given plane.The distance between the line and the parallel plane is the distance between any point on the line and the given plane. Since the plane passes through the point a=- i^ + j^ + k,^the perpendicular distance from the given line to the plane isd=a. n-dn=- i^ + j^ + k^. i ^+ 2 j^ - k^ - 1i^ + 2 j^-k^=-1 + 2 -1-11 + 4 + 1=  16 units

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Question 17:

Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.

Answer:

The equation of the plane passing through the intersection of the given planes is3x-4y+5z-10 + λ 2x+2y-3z-4 = 03+2λ x+-4+2λ y+5-3λ z-10-4λ = 0 ... 1The given line isx = 2y = 3zDividing this equation by 6, we getx6=y3=z2The direction ratios of this line are proportional to 6, 3, 2.So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to 6, 3, 2.3+2λ 6 + -4 + 2λ 3 + 5 - 3λ 2 = 018 + 12λ - 12 + 6λ + 10 - 6λ = 012λ + 16 = 0λ =-4  3Substituting this in (1), we get3+2-4  3 x + -4+2 -4 3 y + 5-3 -4  3 z - 10 - 4 -4  3 = 0x - 20y + 27z = 14



Page No 28.62:

Question 18:

Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines

r=i^+2j^-4k^+λ2i^+3j^+6k^ and r=i^-3j^+5k^+μi^+j^-k^. Also, find the distance of the point (9, −8, −10) from the plane thus obtained.      [CBSE 2014]

Answer:


The equations of the given lines are

r=i^+2j^-4k^+λ2i^+3j^+6k^

r=i^-3j^+5k^+μi^+j^-k^

We know that the vector equation of a plane passing through a point a and parallel to b and c is given by r-a.b×c=0.

Here, a=i^+2j^-4k^, b=2i^+3j^+6k^ and c=i^+j^-k^.

b×c=i^j^k^23611-1=-9i^+8j^-k^

So, the vector equation of the plane is

r-a.b×c=0r-i^+2j^-4k^.-9i^+8j^-k^=0r.-9i^+8j^-k^=i^+2j^-4k^.-9i^+8j^-k^r.-9i^+8j^-k^=1×-9+2×8+-4×-1=-9+16+4=11

Thus, the vector equation of the plane is r.-9i^+8j^-k^=11.

The Cartesian equation of this plane is

xi^+yj^+zk^.-9i^+8j^-k^=11-9x+8y-z=11

Now,

Distance of the point (9, −8, −10) from the plane -9x+8y-z=11

= Length of perpendicular from (9, −8, −10) from the plane -9x+8y-z-11=0

=-9×9+8×-8--10-11-92+82+-12=-81-64+10-1181+64+1=-146146=146 units

Page No 28.62:

Question 19:

Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line x+32=y-37=z-25.

Answer:

The equation of the plane passing through the point (3,4, 1) isa x-3+b y-4+c z-1=0 ... 1This plane passes through (0, 1, 0). So,a 0-3 + b 1-4 + c 0-1 = 0-3a - 3b - c = 0 3a + 3b + c = 0  ... 2Again plane (1) is parallel to the given line. It means that the normal to plane (1) is perpendicular to the line.a 2+b 7+c 5=0  ... 3 (Because a1a2+b1b2+c1c2=0)Solving (1), (2) and (3), we getx-3y-4z-1331275 = 08 x - 3 - 13 y - 4 + 15 z - 1 = 08x - 13y + 15z + 13 = 0

Page No 28.62:

Question 20:

Find the coordinates of the point where the line x-23=y+14=z-22 intersects the plane xy + z − 5 = 0. Also, find the angle between the line and the plane.

Answer:

Let x-23 = y+14 = z-22 = λ (say)x = 3λ + 2;  y = 4λ - 1;  z = 2λ + 2 ... 1Since (xyz) intersects the plane x - y + z - 5 = 0,3λ + 2 - 4λ - 1 + 2λ + 2 - 5 = 03λ + 2 - 4λ + 1 + 2λ + 2 - 5 = 0λ = 0Substituting this in (1), we getx = 2; y = -1;  z = 2So, x, y, z = 2, -1, 2Finding the angleThe given line is parallel to the vector b=3 i^+4 j^+2 k^ and the given plane is normal to the vector n=i^-j^+k^.We know that the angle θ between a line and a plane is given bysin θ=b. nb n=3 i^+4 j^+2 k^. i^-j^+k^3 i^+4 j^+2 k^ i^-j^+k^ = 3 - 4 + 29 + 16 + 4 1 + 1 + 1 = 187θ=sin-1 187

Page No 28.62:

Question 21:

Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane r·i^+2j^-5k^+9=0.

Answer:

Let a, b, c be the direction ratios of the given line.Since the line passes through the point (1, 2, 3) is,x-1a = y-2b = z-3c ...1Since this line is perpendicular to the planeri^ + 2 j^ - 5 k^ + 9 = 0  or  x + 2y - 5z + 9 = 0, the line is parallel to the normal of the plane.So, the direction ratios of the line are proportional to the direction ratios of the given plane.So, a1 = b2 = c-5 = λa = λ; b = 2λ;  c = -5λSubstituting these values in (1), we getx-11 = y+12 = z-2-5, which is the Cartesian form of the line.Vector formThe given line passes through a point whose position vector is a = i^ + 2 j ^+ 3 k^ and is parallel to the vector b = i^ + 2 j^ - 5 k^. So, its equation in vector form isr = a + λbr = i^ + 2 j^ + 3 k^  + λ i ^+ 2 j^ - 5 k^

Page No 28.62:

Question 22:

Find the angle between the line x+12=y3=z-36 and the plane 10x + 2y − 11z = 3.

Answer:

The given line is parallel to the vector b=2 i^+3 j^+6 k^ and the given plane is normal to the vector n=10 i^+2 j^-11 k^.We know that the angle θ between the line and the plane  is given bysin θ=b. nb n=2 i^+3 j^+6 k^. 10 i^+2 j^-11 k^2 i^+3 j^+6 k^ 10 i^+2 j^-11 k^ = 20 + 6 - 664 + 9 + 36 100 + 4 + 121 = -407 15 = -821θ = sin-1 -821

Page No 28.62:

Question 23:

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes r·i^-j^+2k^=5 and r·3i^+j^+k^=6.

Answer:

Let the direction ratios of the required line be proportional to a, b, c. It passes through (1, 2, 3). So its equations arex-1a = y-2b = z-3c ... 1It is given that (1) is parallel to the planes ri^-j^+2 k^ = 5 and r3 i^+j^+2 k^ = 6 or x-y+2z=5 and 3x+y+2z=6Thus,a-b+2c = 0 ... 2 3a + b + z = 0 ... 3Solving these two by cross-multiplication method, we geta-1-2 = b6-1 = c1+3a-3 = b5 = c4 = λ(say)a = -3λ;  b = 5λ; c = 4λSubstituting these values in (1), we getx-1-3 = y-25 = z-34, which is the Cartesian form of the required line.

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Question 24:

Find the value of λ such that the line x-26=y-1λ=z+5-4 is perpendicular to the plane 3xy − 2z = 7.

Answer:

Direction ratios of the given line are proportional to 6, λ, -4.Direction ratios of the plane are 3, -1, -2.Since the given line is parallel to the given plane, the line is perpendicular to the normal of the given plane.6 3+λ -1+-4 -2=018-λ+8=0λ=26

Disclaimer: It should be "parallel" instead of "perpendicular" in the given problem.

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Question 25:

Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line x-11=2y+12=z+1-1.        [CBSE 2015]

Answer:

The general equation of the plane passing through the point (−1, 2, 0) is given by

ax+1+by-2+cz-0=0                 .....(1)

If this plane passes through the point (2, 2, −1), we have

a2+1+b2-2+c-1-0=03a-c=0                                           .....2

Direction ratio's of the normal to the plane (1) are a, b, c.

The equation of the given line is x-11=2y+12=z+1-1. This can be re-written as
x-11=y+121=z+1-1

Direction ratio's of the line are 1, 1, −1.

The required plane is parallel to the given line when the normal to this plane is perpendicular to this line.

a×1+b×1+c×-1=0a+b-c=0                                  .....3

Solving (2) and (3), we get

a0+1=b-1+3=c3-0a1=b2=c3=λSaya=λ,b=2λ,c=3λ

Putting these values of a, b, c in (1), we have

λx+1+2λy-2+3λz-0=0x+1+2y-4+3z=0x+2y+3z=3

Thus, the equation of the required plane is x + 2y + 3z = 3.



Page No 28.65:

Question 1:

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the
(i) yz-plane
(ii) zx-plane

Answer:

i The equation of the line through the points (5, 1, 6) and (3, 4, 1) isx-53-5 = y-14-1 = z-61-6x-5 -2 = y-13 = z-6-5The coordinates of any point on this line are of the formx-5-2 = y-13 = z-6-5 = λx = -2λ+5;  y = 3λ+1; z = -5λ+6So, the coordinates of the point on the given line are -2λ+5, 3λ+1, -5λ+6Since this point lies on the YZ- plane, x = 0-2λ + 5 = 0λ = 52So, the coordinates of the point are-2λ+5, 3λ+1, -5λ+6=-2 52+5, 3 52+1, -5 52+6=0, 172, 132

ii The equation of the line through the points (5, 1, 6) and (3, 4, 1) isx-53-5=y-14-1=z-61-6x-5-2=y-13=z-6-5The coordinates of any point on this line are of the formx-5-2=y-13=z-6-5=λx=-2λ+5; y=3λ+1; z=-5λ+6So, the coordinates of the point on the given line are -2λ+5, 3λ+1, -5λ+6. Since this point lies on the ZX - plane, y=03λ+1=0λ=-13So, the coordinates of the point are-2λ+5, 3λ+1, -5λ+6=-2 -13+5, 3 -13+1, -5 -13+6=173, 0, 233

Page No 28.65:

Question 2:

Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.

Answer:

The equation of the line through the points (3, -4, -5) and (2, -3, 1) isx-32-3 = y+4-3+4 = z+51+5x-3-1 = y+41 = z+56The coordinates of any point on this line are of the formx-3-1 = y+41 = z+56 = λx = -λ+3;  y = λ-4;  z = 6λ-5So, the coordinates of the point on the given line are -λ+3, λ-4, 6λ-5Since this point lies on the plane 2x + y + z = 7,2 -λ+3 + λ-4 + 6λ-5 = 7-2λ + 6 + λ - 4 + 6λ - 5 = 7 5λ = 10   λ = 2So, the coordinates of the point are-λ+3, λ-4, 6λ-5=-2+3, 2-4, 6 2-5=1, -2, 7

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Question 3:

Find the distance of the point (−1, −5, −10) from the point of intersection of the line r=2i^-j^+2k^+λ3i^+4j^+2k^ and the plane r.i^-j^+k^=5.

Answer:

The given equation of the line isr=2 i^-j^+2 k^+λ 3 i^+4 j^+2 k^r=2+3λ i^+-1+4λ j^+2+2λ k^The coordinates of any point on this line are of the form 2 + 3λ i^ + -1 + 4λ j^ + 2 + 2λ k^. or 2 + 3λ, -1 + 4λ, 2 + 2λSince this point lies on the plane ri^ - j ^+ k^ = 5, 2+3λ i^+-1+4λ j^+2+2λ k^. i^-j^+k^ = 52+3λ+1-4λ+2+2λ-5=0λ=0So, the coordinates of the point are2+3λ, -1+4λ, 2+2λ=2+0, -1+0, 2+0=2, -1, 2Distance between (2, -1, 2) and (-1, -5, -10)=-1-22+-5+12+-10-22=9+16+144=13 units

Page No 28.65:

Question 4:

Find the distance of the point (2, 12, 5) from the point of intersection of the line r=2i^-4j^+2k^+λ3i^+4j^+2k^ and r.i^-2j^+k^=0.     [CBSE 2014]

Answer:


The equation of the given line is r=2i^-4j^+2k^+λ3i^+4j^+2k^.

The position vector of any point on the given line is

r=2+3λi^+-4+4λj^+2+2λk^                      .....(1)

If this lies on the plane r.i^-2j^+k^=0, then

2+3λi^+-4+4λj^+2+2λk^.i^-2j^+k^=02+3λ-2-4+4λ+2+2λ=02+3λ+8-8λ+2+2λ=03λ=12λ=4

Putting λ=4 in (1), we get 2+3×4i^+-4+4×4j^+2+2×4k^ or 14i^+12j^+10k^ as the coordinate of the point of intersection of the given line and the plane.

The position vector of the given point is 2i^+12j^+5k^.

∴ Required distance = Distance between 14i^+12j^+10k^ and 2i^+12j^+5k^

                               =14i^+12j^+10k^-2i^+12j^+5k^=12i^+5k^=122+02+52=169=13 units

Page No 28.65:

Question 5:

Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane x-y+z=5.                         [CBSE 2014, 2015]

Answer:


The equation of the line passing through the points A(2, −1, 2) and B(5, 3, 4) is given by

x-25-2=y--13--1=z-24-2Or x-23=y+14=z-22

The coordinates of any point on the line

x-23=y+14=z-22=λsay are 3λ+2,4λ-1,2λ+2         .....(1)

If it lies on the plane x-y+z=5, then

3λ+2-4λ-1+2λ+2=5λ+5=5λ=0

Putting λ=0 in (1), we get (2, −1, 2) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (−1, −5, −10) and (2, −1, 2)

                               =2+12+-1+52+2+102=9+16+144=169=13 units

Page No 28.65:

Question 6:

Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7.                [CBSE 2015]

Answer:


The equation of the line passing through the points A(3, −4, −5) and B(2, −3, 1) is given by

x-32-3=y--4-3--4=z--51--5Or x-3-1=y+41=z+56

The coordinates of any point on the line

x-3-1=y+41=z+56=λsay are -λ+3,λ-4,6λ-5         .....(1)

If it lies on the plane 2x + y + z = 7, then

2-λ+3+λ-4+6λ-5=75λ-3=75λ=10λ=2

Putting λ=2 in (1), we get (1, −2, 7) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (3, 4, 4) and (1, −2, 7)

                               =3-12+4+22+4-72=4+36+9=49=7 units

Page No 28.65:

Question 7:

Find the distance of the point (1, -5, 9) from the plane x-y+z=5 measured along the line x=y=z.

Answer:


The equation of line parallel to the line xyz and passing through the point (1, −5, 9) is

x-11=y+51=z-91     .....(1)

Any point on this line is of the form (k + 1, k − 5, k + 9).

If (k + 1, k − 5, k + 9) be the point of intersection of line (1) and the given plane, then

(k + 1) − (k − 5) + (k + 9) = 5

⇒ k = −10

So, the point of intersection of line (1) and the given plane is (−10 + 1, −10 − 5, −10 + 9) i.e. (−9, −15, −1).

∴ Required distance = Distance between (1, −5, 9) and (−9, −15, −1) = 1+92+-5+152+9+12=3×102=103 units



Page No 28.7:

Question 1:

Write the equation of the plane whose intercepts on the coordinate axes are 2, −3 and 4.

Answer:

We know that the equation of the plane with a, b and c intercepts on the coordianate axes is given by xa+yb+zc=1Given thata = 2;  b = -3;  c = 4So, the equation of the required plane isx2+y-3+z4 = 1 6x - 4y + 3z = 12

Page No 28.7:

Question 2:

Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.
(i) 4x + 3y − 6z − 12 = 0
(ii) 2x + 3y − z = 6
(iii) 2xy + z = 5

Answer:

i Equation of the given plane is 4x+3y-6z-12=04x+3y-6z=12Dividng both sides by 12, we get 4x12 + 3y12 + (-6z)12 = 12124x12 + 3y12  -  6z12 = 1212x3 + y4 + z-2 = 1 ... 1We know that the equation of the plane whose intercepts on the coordianate axes are a, b and is  xa + yb + zc = 1 ... 2Comparing (1) and (2), we geta = 3; b = 4; c =-2

ii The equation of the given plane is       2x + 3y - z = 6    Dividng both sides by 6, we get 2x6 + 3y6 - z6 = 66x3 + y2 + z-6 = 1 ... 1We know that the equation of the plane whose intercepts on the coordianate axes are a, b and is  xa +yb + zc = 1 ... 2Comparing (1) and (2), we geta = 3; b = 2; c =-6

iii  Equation of the given plane is2x-y+z=5Dividng both sides by 5, we get2x5+-y5+z5=55x52+y-5+z5=1 ... 1We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c isxa+yb+zc=1 ... 2Comparing (1) and (2), we geta=52; b=-5; c=5

Page No 28.7:

Question 3:

Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point (α, β, γ).

Answer:

Let ab and c be the intercepts of the given plane on the coordinate axes.Then the plane meets the coordinate axes atA a, 0, 0, B 0, b, 0 and C 0, 0, cGiven that the centroid of the triangle = α, β,γa+0+03, 0+b+03, 0+0+c3=α, β,γa3, b3, c3=α, β,γa3 = α, b3 = β, c3 = γa=3α, b=3β, c=3γ ... 1The equation of the plane whose intercepts on the coordinate axes are ab and c  arexa+yb+zc=1x3α+y3β+z3γ=1 [From (1)]xα+yβ+zγ=3

Page No 28.7:

Question 4:

Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes.

Answer:

We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c isxa+yb+zc=1Given that the plane makes equal intercepts on the coordiante axes.So, a = b = cSo, the equation of the plane isxa+ya+za=1 x + y + z = a ... 1This plane passes through the point (2, 4, 6).Substituting this point in (1), we get2 + 4 + 6 = a a = 12Substituting this in (1), we get x + y + z = 12

Page No 28.7:

Question 5:

A plane meets the coordinate axes at A, B and C, respectively, such that the centroid of triangle ABC is (1, −2, 3). Find the equation of the plane.

Answer:

Let ab and c be the intercepts of the given plane on the coordinate axes.Then the plane meets the coordinate axes atA a, 0, 0, B 0, b, 0 and C 0, 0, cGiven that the centroid of the triangle is 1, -2, 3a+0+03, 0+b+03, 0+0+c3 = 1, -2, 3a3, b3, c3 = 1, -2, 3a3 = 1, b3 = -2, c3 = 3a = 3, b =-6, c = 9 ... 1Equation of the plane whose intercepts on the coordinate axes are ab and c isxa + yb + zc = 1x3 + y-6 + z9 = 1 [From (1)]6x - 3y + 2x = 18



Page No 28.74:

Question 1:

Show that the lines r=2j^-3k^+λi^+2j^+3k^ and r=2i^+6j^+3k^+μ2i^+3j^+4k^ are coplanar. Also, find the equation of the plane containing them.

Answer:

We know that the lines r=a1+λ b1 and r=a2+μ b2 are coplanar if a1. b1×b2 = a2. b1×b2 and the equation of the plane containing them isr. b1×b2 = a1. b1×b2.Here,a1 = 0 i^+2 j^-3 k^;  b1 = i^+2 j^+3 k^;  a2 = 2 i^+6 j^+3 k^;  b2 = 2 i^+3 j^+4 k^b1×b2 = i^j^k^123234 = -i^+2 j^-k^a1. b1×b2 = 0 i^+2 j^-3 k^. -i^+2 j^-k^ = 0 + 4 + 3 = 7a2. b1×b2 = 2 i^ + 6 j^ + 3 k^. -i^ + 2 j ^- k^ = -2 + 12 - 3 = 7Clearly, a1. b1×b2=a2. b1×b2Hence, the given lines are coplanar.The equation of the plane containing the given lines isr. b1×b2 = a1. b1×b2r. -i^+2 j^-k^ = 0 i^ + 2 j^ - 3 k^. -i ^+ 2 j^ - k^r. -i^ + 2 j^ - k^ = 7r. i^ - 2 j^ + k^ + 7 = 0

Page No 28.74:

Question 2:

Show that the lines x+1-3=y-32=z+21 and x1=y-7-3=z+72 are coplanar. Also, find the equation of the plane containing them.

Answer:

We know that the linesx-x1l1=y-y1m1=z-z1n1 and x-x2l2=y-y2m2=z-z2n2 are coplanar ifx2-x1y2-y1z2-z1l1m1n1l2m2n2 = 0and the equation of the plane containing these lines isx-x1y-y1z-z1l1m1n1l2m2n2 = 0Here,x1=-1; y1=3; z1=-2; x2= 0; y2 = 7; z2=-7; l1=-3; m1 = 2; n1=1; l2=1; m2=-3; n2=2Now,0+17-3-7+2-3211-32=14-5-3211-32=1 7 - 4 -7 - 5 7=7 + 28 - 35=0So, the given lines are coplanar.The equation of the plane containing the given lines isx+1y-3z+2-3211-32=0x+1 7 - y-3 -7 + z+2 7 = 07x + 7y + 7z = 0x + y + z = 0



Page No 28.75:

Question 3:

Find the equation of the plane containing the line x+1-3=y-32=z+21 and the point (0, 7, −7) and show that the line x1=y-7-3=z+72 also lies in the same plane.

Answer:

Let the equation of the plane passing through (0, 7, -7) bea x-0 + b y-7 +c  z+7 = 0 ... 1The line x+1-3 = y-32 = z+21 passes through (-1, 3, -2) and its direction ratios are proportional to -3, 2, 1. Since plane (1) contains this line, it must pass through the point (-1, 3, -2).a -1-0 + b 3-7+c -2+7 = 0 -a -4b + 5c = 0a + 4b - 5c = 0... 2Since plane (1) contains this line, it must be parallel to the line.-3a+2b+c=0 ... 3Solving (1), (2) and (3), we getx-0y-7z+714-5-321 = 014x + 14 y - 7 + 14 z + 7 =014x + 14y + 14z = 0x + y + z = 0

Page No 28.75:

Question 4:

Find the equation of the plane which contains two parallel lines x-41=y-3-4=z-25 and x-31=y+2-4=z5.

Answer:

We know that the equation of the plane containing two given parallel lines x-x1a = y-y1b = z-z1c  and  x-x2a = y-y2b = z-z2c isx-x1y-y1z-z1x2-x1y2-y1z2-z1a b c=0Here,x1 = 4; y1 = 3;  z1 = 2; x2 = 3;  y2 = -2;  z2 = 0;  l1 = 1;  m1 = -4;  n1 = 5;  l2 = 1;  m2 = -4;  n2 = 5Now,x-4y-3z-23-4-2-30-21-45 = 0x-4y-3z-23-4-2-3 0-21-45 = 0-33 x-4 + 3 y-3 + 9 z-2 = 011 x-4 - y-3 - 3 z-2 = 011x-y-3z = 35

Page No 28.75:

Question 5:

Show that the lines x+43=y+65=z-1-2 and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.

Answer:

The equation of the given line isx+43 = y+65 = z-1-2The coordinates of any point on this line are of the formx+43 = y+65 = z-1-2 = λx = 3λ-4; y = 5λ-6; z = -2λ+1So, the coordinates of the point on the given line are 3λ-4, 5λ-6, -2λ+1. Since this point lies on the plane       3x - 2y + z + 5 = 0,3 3λ - 4 - 2 5λ - 6 + -2λ + 1 + 5 = 09λ - 12 - 10λ + 12 - 2λ + 1 + 5 = 0-3λ + 6 = 0λ = 2So, the coordinates of the point are3λ - 4, 5λ - 6, -2λ + 1=3 2 - 4, 5 2 - 6, -2 2 + 1=2, 4, -3Substituting this point in another plane equation 2x+3y+4z-4=0, we get2 2+3 4+4 -3-4=04+12-12-4=00=0So, the point (2, 4, -3) lies on another plane too. So, this is the point of intersection of both the lines.Finding the plane equationLet the direction ratios be proportional to a, b, c.Since the plane contains the line x+43= y+65 = z-1-2, it must pass through the point (-4, -6, 1) and is parallel to this line.So, the equation of plane isa x + 4 + b y + 6 + c z - 1 = 0 ...1and3a + 5b - 2c = 0... 2Since the given plane contains the planes 3x - 2y + z + 5 = 0 = 2x + 3y + 4z - 4, 3a - 2b + c = 0 ... 32a + 3b + 4z = 0 ... 4Solving (3) and (4) using cross-multiplication, we geta-11 = b-10 = c13... 5Using (1), (2) and (5), the equation of plane is x+4 y+6 z-135-21110-13 = 0-45 x+4+17 y+6-25 z-1=045 x+4-17 y+6+25 z-1=045x-17y+25z+53=0 

Page No 28.75:

Question 6:

Show that the plane whose vector equation is r·i^+2j^-k^=3 contains the line whose vector equation is r=i^+j^+λ2i^+j^+4k^.

Answer:

The line  r=i^+j^+0 k^+λ 2 i^+j^+4 k^....(1) passes through a point whose position vector is a=i^+j^+0 k^ and is parallel to the vector b=2 i^+j^+4 k^.If the plane ri^+2 j^-k^=3 contains the given line, then (1) it should passes through the point i^+j^+0 k^ (2) it should be parallel to the lineNow, i^+j^+0 k^i^+2 j^-k^ = 1 + 2 = 3So, the plane passes through the point i^+j^+0 k^.The normal vector to the given plane is n =  i^+ 2 j^ - k.^We observe thatb. n = 2 i^ + j ^+ 4 k^. i^ + 2 j^ - k^ = 2 + 2 - 4 = 0Therefore, the plane is parallel to the line.Hence, the given plane contains the given line.

Page No 28.75:

Question 7:

Find the equation of the plane determined by the intersection of the lines x+33=y-2=z-76 and x+61=y+5-3=z-12

Answer:

  The given equations of the lines arex+33 = y-2 = z-76... 1x+61 = y+5-3 = z-12... 2Let the direction ratios of the plane be proportional to a, b, c.Since the plane contains line (1), it should pass through (-3, 0, 7) and is parallel to the line (1).Equation of the plane through (1) isa x + 3 + b y + c z - 7 = 0 ... 3,where 3a - 2b + 6c = 0 ... 4Since the plane contains line (2), the plane is parallel to line (2) also.a - 3b + 2c = 0 ... 5Solving (4) and (5) using cross-multiplication, we geta14 = b0 = c-7Substituting a, b and in (3), we get14 x + 3 + 0 y - 7 z - 7 = 02 x + 3 + 0 y - 1 z - 7 = 02x - z + 13 = 0

Page No 28.75:

Question 8:

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line r=i^+3j^-2k^+λi^-j^+k^.

Answer:

The equation of any plane passing through (3, 4, 2) isa x-3 + b y-4 + c z-2 = 0 ... 1It is given that (1) is  passing through (7, 0, 6). So,a 7-3 + b 0-4 + c 6-2 = 0 4a - 4b + 4c = 0a - b + c = 0 ... 2It is given that (1) is perpendicular to the plane 2x - 5y + 0z + 15z = 0. So,2a - 5b + 0c = 0 ... 3Solving (1), (2) and (3), we getx-3y-4z-21-112-50=05 x-3 + 2 y-4 -3 z-2 = 05x + 2y - 3z = 17Or r5 i^ + 2 j^ - 3 k^  = 17Showing that the plane contains the lineThe line  r=i^+3 j^-2 k^+λ i^-j^+k^ passes through a point whose positon vector is a = i^ + 3 j^-2 k^ and is parallel to the vector b = i ^- j ^+ k^.If the plane r5 i^+2 j^-3 k^=17 contains the given line, then (1) it should pass through the point i^+3 j^-2 k^ (2) it should be parallel to the lineNow, i^+3 j^-2 k^5 i^+2 j^-3 k^= 5 + 6 + 6 = 17So, the plane passes through the point i^ + 3 j ^- 2 k^.The normal vector to the given plane is n =  i ^- j^ + k.^We observe thatb. n=i^-j^+k^. 5 i^+2 j^-3 k^ = 5 - 2 - 3 = 0Therefore, the plane is parallel to the line.Hence, the given plane contains the given line.

Page No 28.75:

Question 9:

If the lines x-1-3=y-2-2k=z-32 and x-1k=y-21=z-35 are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.

Answer:

We know that the linesx-x1l1 = y-y1m1 = z-z1n1 and x-x2l2 = y-y2m2 = z-z2n2 are perpendicular ifl1l2 + m1m2 + n1n2 = 0Here,l1 = -3; m1=-2k; n1= 2; l2 = k; m2 = 1; n2 = 5It is given that given lines are perpendicular.l1l2+m1m2+n1n2 = 0-3 k + -2k 1 + 2 5 = 0-3k - 2k + 10 = 0-5k = -10       k = 2Substituting this value in the given equations of the lines, we getx-1-3 = y-2-4 = z-32... 1 x-12 = y-21 = z-35 ... 2Finding the equation of the planeLet the direction ratios of the required plane be proportional to a, b, c.We know from (1) and (2) that lines (1) and (2) pass through the point (1, 2, 3) and the direction ratios of (1) and (2) are proportional to -3, -4, 2 and 2, 1, 5 respectively.Since the plane contains the lines (1) and (2), the plane must pass through the point (1, 2, 3) and it must be parallel to the line.So, the equation of the plane isa x - 1 + b y - 2 + c z - 3 = 0 ... 3-3a - 4b + 2c = 0 ... 42a + b + 5c = 0 ... 5Solving (1), (2) and (3), we getx-1y-2z-3-3-42215=0-22 x-1 + 19 y-2 + 5 z-3 = 0-22x + 19y + 5z = 31

Page No 28.75:

Question 10:

Find the coordinates of the point where the line x-23=y+14=z-22 intersect the plane xy + z − 5 = 0. Also, find the angle between the line and the plane.

Answer:

The coordinates of any point on this line are of the formx-23=y+14=z-22=λx = 3λ + 2; y = 4λ - 1; z = 2λ + 2So, the coordinates of the point on the given line are 3λ + 2, 4λ - 1, 2λ + 2. This point lies on the plane x - y + z - 5 = 0.3λ + 2 - 4λ + 1 + 2λ + 2 - 5 = 0λ = 0So, the coordinates of the point are3λ + 2, 4λ - 1, 2λ + 2=3 0 + 2, 4 0 - 1, 2 0 + 2=2, -1, 2Finding the angle between the line and the planeThe given line is parallel to the vector b=3 i^+4 j^+2 k^ and the given plane is normal to the vector n=i ^- j^ + k^.We know that the angle θ between the line and the plane  is given bysin θ=b. nb n=3 i^ + 4 j^ + 2 k^. i^ - j^ + k^3 i^ + 4 j^ + 2 k^ i^ - j ^+ k^ = 3 - 4 + 29 + 16 + 4 1 + 1 +1 = 187θ = sin-1 187

Page No 28.75:

Question 11:

Find the vector equation of the plane passing through three points with position vectors i^+j^-2k^, 2i^-j^+k^ and i^+2j^+k^. Also, find the coordinates of the point of intersection of this plane and the line r=3i^-j^-k^+λ2i^-2j^+k^.

Answer:

Let A (1, 1 , -2), B (2, -1, 1) and C (1, 2, 1) be the points represented by the given position vectors.The required plane passes through the point  A (1, 1, -1) whose position vector is a=i^ +  j^ - 2 k^ and is normal to the vector n given byn=AB × AC.Clearly, AB=OB -OA = 2 i^- j^+k^-i^ +  j^ - 2 k^ = i^-2 j^+3 k^AC=OC-OA=i^+2  j^+k^-i^j^-2 k^=0 i^+ j^+3 k^n=AB×AC=i^j^k^1-23013=-9 i^-3 j^ +k^The vector equation of the required plane isr. n=a. nr. -9 i^-3 j^ +k^=i^j^-2 k^. -9 i^-3 j^ +k^r. -9 i^-3 j^ +k^=-9-3-2r. -9 i^+3 j^ -k^=-14r. 9 i^+3 j^ -k^=14To find the point of intersection of this plane The given equation of the line isr=3 i^-j^-k^+λ 2 i^-2 j^+k^r=3+2λ i^+-1-2λ j^+-1+λ k^The coordinates of any point on this line are in the form of  3 + 2λ i ^+ -1 - 2λ j ^+ -1 + λ k^ or 3 + 2λ, -1 - 2λ, -1 + λSince this point lies on the plane r9 i^ + 3 j^ -k^= 14,3 + 2λ i^ + -1 - 2λ j ^+ -1 + λ k^. 9 i^ + 3 j^ -k^ = 1427 + 18λ - 3 - 6λ + 1 - λ = 1411λ =-11λ =-1So, the coordinates of the point are3 + 2λ, -1 - 2λ, -1 + λ=3 - 2, -1 + 2, -1 - 1=1, 1, -2

Page No 28.75:

Question 12:

Show that the lines 5-x-4=y-74=z+3-5 and x-87=2y-82=z-53 are coplanar.         [CBSE 2014]

Answer:


The equations of the given lines can be re-written as

x-54=y-74=z+3-5

and x-87=y-41=z-53

We know that the lines x-x1l1=y-y1m1=z-z1n1 and x-x2l2=y-y2m2=z-z2n2 are coplanar if x2-x1y2-y1z2-z1l1m1n1l2m2n2=0.

Here,

x1=5,y1=7,z1=-3,x2=8,y2=4,z2=5l1=4,m1=4,n1=-5,l2=7,m2=1,n2=3

x2-x1y2-y1z2-z1l1m1n1l2m2n2=8-54-75--344-5713=3-3844-5713=312+5+312+35+84-28=51+141-192=0

So, the given lines are coplanar.

Page No 28.75:

Question 13:

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line x-31=y-65=z-44.         [CBSE 2015]

Answer:


Let the equation of the plane passing through (3, 2, 0) be

ax-3+by-2+cz-0=0                     .....(1)

The line x-31=y-65=z-44 passes through the point (3, 6, 4) and its direction ratios are proportional to 1, 5, 4.

If plane (1) contains this line, then it must pass through (3, 6, 4) and must be parallel to the line.

a3-3+b6-2+c4-0=04b+4c=0b+c=0                              .....2

Also,

1×a+5×b+4×c=0a+5b+4c=0             .....3

Solving (2) and (3), we get

a4-5=b1-0=c0-1a-1=b1=c-1=λSaya=-λ,b=λ,c=-λ

Putting these values of a, b, c in (1), we get

-λx-3+λy-2-λz-0=0-x+3+y-2-z=0-x+y-z+1=0x-y+z-1=0

Thus, the equation of the required plane is xy + z − 1 = 0.

Page No 28.75:

Question 14:

Show that the lines x+3-3=y-11=z-55 and x+1-1=y-22=z-55 are coplanar. Hence, find the equation of the plane containing these lines.

Answer:


The lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 are coplanar if x2-x1y2-y1z2-z1a1b1c1a2b2c2=0.

The given lines are x+3-3=y-11=z-55 and x+1-1=y-22=z-55.

Now, x2-x1y2-y1z2-z1a1b1c1a2b2c2=-1--32-15-5-315-125=210-315-125=25-10-1-15+5+0=-10+10+0=0

So, the given lines are coplanar.

The equation of the plane containing the given lines is

x--3y-1z-5-315-125=0x+3y-1z-5-315-125=0x+35-10-y-1-15+5+z-5-6+1=0-5x+3+10y-1-5z-5=0x-2y+z=0
Thus, the equation of the plane containing the given lines is x − 2yz = 0.

Page No 28.75:

Question 15:

 If the linex-32=y+2-1=z+43 lies in the plane lx+my-z=9, then find the value of l2+m2.

Answer:


The line x-x1a=y-y1b=z-z1c lies in the plane AxByCz D = 0 iff (i) Ax1+By1+Cz1+D=0 and (ii) aA+bB+cC=0.

It is given that the line x-32=y+2-1=z+43 lies in the plane lx+my-z=9.

l×3+m×-2--4=93l-2m=5         .....1
Also,

2×l+-1×m+3×-1=02l-m=3          .....2
Solving (1) and (2), we get

l = 1 and m = −1

∴ l2m2 = 12 + (−1)2 = 1 + 1 = 2

Thus, the value of l2 + m2 is 2.

Page No 28.75:

Question 16:

Find the values of λ for which the lines x-11=y-22=z+3λ2 and x-31=y-2λ2=z-12 are coplanar.

Answer:


The lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 are coplanar if x2-x1y2-y1z2-z1a1b1c1a2b2c2=0.

The given lines x-11=y-22=z+3λ2 and x-31=y-2λ2=z-12 are coplanar.

x2-x1y2-y1z2-z1a1b1c1a2b2c2=0
3-12-21--312λ21λ22=020412λ21λ22=024-λ4-0+4λ2-2=0-2λ4+4λ2=0
λ2λ2-2=0λ2=0 or λ2-2=0λ=0 or λ=±2
Thus, the values of λ are 0, -2 and 2.

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Question 17:

If the lines x=5,  y3-α=z-2 and x=αy-1=z2-α are coplanar, find the values of α.

Answer:


The lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 are coplanar if x2-x1y2-y1z2-z1a1b1c1a2b2c2=0.

The given lines x-50=y3-α=z-2 and x-α0=y-1=z2-α are coplanar.

 α-50-00-003-α-20-12-α=0α-50003-α-20-12-α=0α-53-α×2-α-2-0+0=0α-5α2-5α+4=0α-5α-1α-4=0
α-1=0 or α-4=0 or α-5=0α=1 or α=4 or α=5
Thus, the values of α are 1, 4 and 5.

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Question 18:

If the straight lines x-12=y+1k=z2 and x+12=y+12=zk are coplanar, find the equations of the planes containing them.

Answer:


The lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 are coplanar if x2-x1y2-y1z2-z1a1b1c1a2b2c2=0.

The given lines  x-12=y+1k=z2 and x+12=y+12=zk are coplanar.

 -1-1-1--10-02k222k=0-2002k222k=0-2k2-4-0+0=0k2-4=0k=±2
The equation of the plane containing the given lines is x-1y+1z2k222k=0.

For k = 2,  x-1y+1z2k222k=x-1y+1z222222=0
So, no plane exists for k = 2.

For k = −2,
x-1y+1z2k222k=0
x-1y+1z2-2222-2=0x-14-4-y+1-4-4+z4+4=08y+1+8z=0y+z+1=0
Thus, the equation of the plane containing the given lines is yz + 1 = 0.

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Question 19:

Find the vector equation of the plane that contains the lines r= i^+j^ +λ i^+2j^-k^  and the point (-1, 3, -4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane, thus obtained.

Answer:

Given line r=i^+j^+λi^+2j^-k^ passes through (1,1,0) and is parallel to the vector i^+2j^-k^

i.e. The given plane passes through (1,1,0) and B(−1, 3, −4) and is parallel to b=c+2j^-k^

Let n be the normal vector to the required plane. 

Then n is perpendicular to both b and AB 

i.e. n is parallel to AB×B
Let n1=AB×bthen n1=i^j^k^-22-412-1            =i^-2+8-j^2+4+R-4-2     n1=6i^-6j^-6k^Let α be position vector of A i.e. α=i^+j^

Then the required plane passes through α=i^+j^ and is perpendicular to n1=6i^-6j^-6k^
So, its vector equation is 
r-α.n1=0i.e. r.n1=α.n1i.e. r.6i^-6j^-6k^=i^+j^.6i^-6j^-6k^r.6i^-6j^-6k^=6-6=0i.e. r.i^-j^-k^=0for r=xi^+yj^+zk^
Equation of plane is x – y – z = 0
The length of perpendicular form (2, 1, 4) to plane x – y – z = 0 is
d=2i^+j^+4k^.i^-j^-k^12+-12+-12=2-1-43  =33=3 i.e. d=3



Page No 28.78:

Question 1:

Find the shortest distance between the lines x-2-1=y-52=z-03 and x-02=y+1-1=z-12.

Answer:

The given equations of the lines arex-2-1 = y-52 = z-03... 1x-02 = y+1-1 = z-12... 2Clearly (2) passes through the point P (0, -1, 1).Let the direction ratios of the plane be proportional to a, b, c.Since the plane containing line (1) should pass through (2, 5, 0) and is parallel to the line (1),equation of the plane passing through (1) isa x-2 + b y-5 + c z-0 = 0 ... 3,where -a + 2b + 3c = 0 ... 4Since the plane is parallel to line (2),2a - b + 2c = 0 ... 5Solving (4) and (5) using cross-multiplication, we geta7 = b8 =  c-3Substituting a, b and in (3), we get7 x-2 + 8 y-5 - 3 z-0 = 07x + 8y - 3z - 54 = 0 ... 6,which is the equation of the plane containing line (1) and parallel to line (2).Shortest distance between (1) and (2)=Distance between the point P (0, -1, 1) and plane (6)=7 0+8 -1-3 1-5449+64+9=65122 units

Page No 28.78:

Question 2:

Find the shortest distance between the lines x+17=y+1-6=z+11 and x-31=y-5-2=z-71.

Answer:

The given equations of the lines arex+17 = y+1-6 = z+11... 1x-31 = y-5-2 = z-71... 2Clearly (2) passes through the point P (3, 5, 7).Let the direction ratios of the plane be proportional to a, b, c.Since the plane contains line (1), it should pass through (-1, -1, -1) and is parallel to the line (1).Equation of the plane through (1) isa x+1+b y+1+c z+1 = 0 ... 3,where 7a - 6b + c = 0 ... 4Since the plane is parallel to the line (2),a-2b+c=0 ... 5Solving (4) and (5) using cross-multiplication, we geta-4=b-6=c-8a2=b3=c4Substituting a, b and in (3), we get2 x+1+3 y+1+4 z+1=02x+3y+4z+9=0 ... 6which is the equation of the plane containing line (1) and parallel to line (2).Shortest distance between (1) and (2)=Distance between the point P (3, 5, 7) and plane (6)=2 3+3 5+4 7+94+9+16=5829=2 29 units

Page No 28.78:

Question 3:

Find the shortest distance between the lines x-12=y-34=z+21 and 3x-y-2z+4=0=2x+y+z+1.

Answer:


The equation of the plane containing the line 3x-y-2z+4=0=2x+y+z+1 is

3x-y-2z+4+λ2x+y+z+1=0Or 3+2λx+λ-1y+λ-2z+λ+4=0                 .....1

If it is parallel to the line x-12=y-34=z+21, then

23+2λ+4λ-1+λ-2=09λ=0λ=0

Putting λ=0 in (1), we get

3x-y-2z+4=0                         .....2

This is the equation of the plane containing the second line and parallel to the first line.

Now, the line x-12=y-34=z+21 passes through (1, 3, −2).

∴ Shortest distance between the given lines

= Length of the perpendicular from (1, 3, −2) to the plane 3x-y-2z+4=0

=3×1-3-2×-2+432+-12+-22=3-3+4+49+1+4=814 units



Page No 28.83:

Question 1:

Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.

Answer:

Let Q be the image of the point P (0, 0, 0) in the plane 3x + 4y - 6z + 1 = 0 .Then, PQ is normal to the plane. So, the direction ratios of PQ are proportional to 3, 4, -6.Since PQ passes through P (0, 0, 0) and has direction ratios proportional to 3, 4 and -6, equation of PQ isx-03 = y-04 = z-0-6 = r (say)Let the coordiantes of Q be 3r, 4r, -6r. Let R be the mid-point of PQ. Then,R=0+3r2, 0+4r2, 0-6r2 = 3r2, 2r, -3rSince R lies in the plane 3x + 4y - 6z + 1 = 0,3 3r2 + 4 2r - 6 -3r + 1  = 0r = -261Substituting this in the coordinates of Q, we getQ=3r, 4r, -6r = 3 -261, 4 -261, -6 -261 = -661, -861, 1261

Page No 28.83:

Question 2:

Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.

Answer:

Let Q be the image of the point P (1, 2, -1) in the plane 3x-5y+4z=5.Then PQ is normal to the plane.  So, the direction ratios of PQ are proportional to 3, -5, 4.Since PQ passes through P (1, 2, -1) and has direction ratios proportional to 3, -5, 4, equation of PQ isx-13=y-2-5=z+14=r (say)Let the coordinates of Q be 3r+1, -5r+2, 4r-1. Let R be the mid-point of PQ. Then,R=3r + 1 + 12, -5r + 2 + 22, 4r - 1 - 12 = 3r + 22, -5r + 42,  4r - 22Since R lies in the plane 3x - 5y + 4z = 5,3 3r + 22 -5 -5r + 42 +4 4r - 22 = 59r + 6 + 25r - 20 + 16r - 8 = 1050r - 32 = 0r = 3250 = 1625Substituting the value of r in the coordinates of Q, we getQ=3r + 1, -5r + 2, 4r - 1 = 3 1625+1, -5 1625+2, 4 1625-1 = 7325, -65, 3925

Page No 28.83:

Question 3:

Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line x+12=y-33=z-1-1. Hence, or otherwise, deduce the length of the perpendicular.

Answer:

Let M be the foot of the perpendicular of the point P (5, 4, 2) on the line x + 12 = y - 33 = z - 1-1Therefore, its equation isx + 12 = y - 33 = z - 1-1 = rThen, M is in the form 2r-1, 3r+3, -r+1Direction ratios of MP are 2r-1-5, 3r+3-4, -r+1-2 or 2r-6, 3r-1, -r-1.Since MP is perpendicular to the given line (2, 3, -1),2 2r-6+3 3r-1-1 -r-1=0 (Because a1a2+b1b2+c1c2=0)4r-12+9r-3+r+1=014r-14=0r=1So, M=2r-1, 3r+3, -r+1=2 1-1, 3 1+3, -1+1=1, 6, 0Length of the perperndicular, MP=1-52+6-42+0-22=16+4+4=24=2 6 units

Page No 28.83:

Question 4:

Find the image of the point with position vector 3i^+j^+2k^ in the plane r·2i^-j^+k^=4. Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through 3i^+j^+2k^.

Answer:


Let Q be the image of the point P (3 i^+j^+2 k^) in the plane r. 2 i^-j^+ k^ = 4Since PQ passes through P and is normal to the given plane, it is parallel to the normal vector 2 i^-j^+k^. So, the equation of PQ isr=i^+j^+2 k^+λ 2 i^-j^+k^As Q lies on PQ, let the position vector of Q be 3+2λ i^+1-λ j^+2+λ k^ .Let R be the mid-point of PQ. Then, the position vector of R is 3+2λ i^+1-λ j^+2+λ k^+i^+j^+2 k^2=6+2λ i^+2-λ j^+4+λ k^2=3+λ i^+1-λ2 j^+2+λ2 k^Since R lies in the plane r. 2 i^-j^+ k^ = 4,3+λ i^+1-λ2 j^+2+λ2 k^. 2 i^-j^+ k^ = 46+2λ-1+λ2+2+λ2=47 + 2λ + λ2 + λ2 = 4 14 + 6 λ = 8 6 λ = 8 - 14  λ = -1Putting  λ = -1 in Q, we get Q= 3+2(-1) i^+1-(-1) j^+2+(-1) k^   =  i^ + 2 j^ +k ^or (1, 2, 1)Therefore, by putting  λ = -1 in R, we get R = 3+(-1) i^+1-(-1)2 j^+2+(-1)2 k^= 2 i^ + 32 j^ +32 k^          

Page No 28.83:

Question 5:

Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.

Answer:

Let M be the foot of the perpendicular of the point P (1, 1, 2) in the plane 2x - 2y + 4z + 5 = 0Then, PM is normal to the plane. So, the direction ratios of PM are proportional to 2, -2, 4.Since PM passes through P (1, 1, 2) and has direction ratios proportional to 2, -2 and 4, equation of PQ isx-12 = y-1-2 = z-24 = r (say)Let the coordiantes of M be 2r + 1, -2r + 1, 4r + 2.Since M lies in the plane 2x - 2y + 4z + 5 = 0,2 2r + 1 - 2 -2r + 1 + 4 4r + 2 + 5 =04r + 2 + 4r - 2 + 16r + 8 + 5 = 024r + 13 = 0r = -13 24Substituting this in the coordinates of M, we getM=2r + 1, -2r + 1, 4r + 2 = 2 -13 24 + 1, -2 -13 24 + 1, 4 -13 24 + 2 = -1 12, 25 12, -16Now, the length of the perpendicular from P onto the given plane=2 1-2 1+4 2+54+4+16=1324 units

Page No 28.83:

Question 6:

Find the distance of the point (1, −2, 3) from the plane xy + z = 5 measured along a line parallel to x2=y3=z-6.

Answer:


The given plane is x − y + z = 5. We need to find the distance of the point from the point(1, -2, 3) measured along a parallel line x2=y3=z-6.
Let the line from point be P(1, -2, 3) and meet the plane at point Q.
Direction ratios of the line from the point ​(1, -2, 3) to the given plane will be the same as the given line x2=y3=z-6.
So the equation of the line passing through P and with same direction ratios will be:
x-12 = y+23 = z-3-6 = λx = 2λ + 1, y = 3λ-2, z = -6λ+3
coordinates of any point on the line PQ are x = 2λ + 1, y = 3λ-2, z = -6λ+3.
Now, since Q lies on the plane so it must satisfy the equation of the plane.
that is, ​x − y + z = 5

therefore, 2λ+1 - (3λ - 2) + (-6λ +3) = 5
λ = 17
coordinates of Q are (27+1), (37-2), (-67+3) = 97, -117, 157
using distance formula we have the length of PQ as
PQ = 97-12 + -117+22 + 157-32      =272 + 372 + -672      =449 + 949 + 3649      =4949 = 1

Hence PQ = 1

So, ​the distance of the point (1, −2, 3) from the plane x − y + z = 5  is 1







 

Page No 28.83:

Question 7:

Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3xyz = 7. Also, find the length of the perpendicular.

Answer:

Let M be the foot of the perpendicular of the point P (2, 3, 7) in the plane 3x - y - z = 7.Then, PM is normal to the plane. So, the direction ratios of PM are proportional to 3, -1, -1.Since PM passes through P (2, 3, 7) and has direction ratios proportional to 3, -1 and -1, equation of PQ isx-23 = y-3-1 = z-7-1 = r (say)Let the coordinates of M be 3r + 2, -r + 3, -r + 7.Since M lies in the plane 3x - y - z = 7,3 3r + 2 - -r + 3 - -r + 7 = 79r + 6 + r - 3 + r - 7 = 711r = 11r = 1Substituting this in the coordinates of M, we getM = 3r + 2, -r + 3, -r + 7 = 3 1 + 2, -1 + 3, -1 + 7 = 5, 2, 6Now, the length of the perpendicular from P onto the given plane=3 2-3-7-79+1+1=1111=11 units

Page No 28.83:

Question 8:

Find the image of the point (1, 3, 4) in the plane 2xy + z + 3 = 0.

Answer:

Let Q be the image of the point P (1, 3, 4) in the plane 2x-y+z+3=0.Then PQ is normal to the plane. So, the direction ratios of PQ are proportional to 2, -1, 1.Since PQ passes through P (1, 3, 4) and has direction ratios proportional to 2, -1 and 1, equation of PQ isx-12=y-3-1=z-41=r (say)Let the coordinates of Q be 2r+1, -r+3, r+4. Let R be the mid-point of PQ. Then,R=2r+1+12, -r+3+32, r+4+42=r+1, -r+62, r+82Since R lies in the plane 2x - y + z + 3 = 0,2 r+1 - -r+62 + r+82 + 3 = 04r + 4 + r - 6 + r + 8 + 6 = 06r + 12 = 0r = -2Substituting this in the coordinates of Q, we getQ=2r+1, -r+3, r+4.=2 -2+1, 2+3, -2+4=-3, 5, 2

Page No 28.83:

Question 9:

Find the distance of the point with position vector -i^-5j^-10k^ from the point of intersection of the line r=2i^-j^+2k^+λ3i^+4j^+12k^ with the plane r·i^-j^+k^=5.

Answer:

The given equation of the line isr=2 i^-j^+2 k^+λ 3 i^+4 j^+2 k^r=2+3λ i^+-1+4λ j^+2+2λ k^The coordinates of any point on this line are of the form 2+3λ i^+-1+4λ j^+2+2λ k^ or 2+3λ, -1+4λ, 2+2λSince this point lies on the plane ri^-j^+k^ = 5, 2+3λ i^+-1+4λ j^+2+2λ k^. i^-j^+k^=52+3λ+1-4λ+2+2λ-5=0λ=0So, the coordinates of the point are2 + 3λ, -1 + 4λ,  2 + 2λ=2 + 0, -1 + 0,  2 + 0=2, -1,  2The coordinates of the point corresponding to the position vector -i^-5 j^-10 k^ are (-1, -5, -10).Distance between (2, -1, 2) and (-1, -5, -10)=-1-22+-5+12+-10-22=9+16+144=13 units

Page No 28.83:

Question 10:

Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane r·i^-2j^+4k^+5=0.

Answer:

Let M be the foot of the perpendicular of the point P (1, 1, 2) in the plane ri^ - 2 j^+ 4 k^ + 5 = 0 or x-2y + 4z + 5 = 0.Then, PM is the normal to the plane.  So, the direction ratios of PM are proportional to 1, -2, 4.Since PM passes through P (1, 1, 2) and has direction ratios proportional to 1,-2, 4 equation of PQ isx-11 = y-1-2 = z-24 = r (say)Let the coordinates of M be r+1, -2r+1, 4r+2.Since M lies in the plane x - 2y + 4z + 5 = 0,x - 2y + 4z + 5 = 0r + 1 + 4r - 2 + 16r + 8 + 5 = 021r + 12 = 0r=-1221=-47Substituting this in the coordinates of M, we getM=r+1, -2r+1, 4r+2= -47+1, -2 -47+1,  4 -47+2 = 37, 157,-27 Now, the length of the perpendicular from P onto the given plane= 1-2 1+4 2+51+4+16=1221 units

Disclaimer: The answer given for this problem in the text book is incorrect.

Page No 28.83:

Question 11:

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the  point P (3, 2, 1) from the plane 2xy + z + 1 = 0. Also, find the image of the point in the plane.

Answer:

Let M be the foot of the perpendicular of the point P (3, 2, 1) in the plane 2x - y + z + 1 = 0.Then, PM is the normal to the plane. So, the direction ratios of PM  are proportional to 2, -1, 1.Since PM passes through P (3, 2, 1) and has direction ratios proportional to 2, -11,equation of PQ isx-32=y-2-1=z-11=r (say)Let the coordinates of M be 2r+3, -r+2, r+1.Since M lies in the plane 2x-y+z+1=0,2 2r + 3 - -r + 2 + r + 1 + 1 = 04r + 6 + r - 2 + r + 2 = 06r=-6r=-1Substituting the value of r in the coordinates of M, we getM=2r + 3, -r + 2, r + 1 = 2 -1 + 3, --1 + 2, -1 + 1 = 1, 3, 0Now, the length of the perpendicular from P onto the given plane=2 3 - 2 + 1 + 14 + 1 + 1= 66=6 units

Page No 28.83:

Question 12:

Find the direction cosines of the unit vector perpendicular to the plane r·6i^-3j^-2k^+1=0 passing through the origin.

Answer:

For the unit vector perpendicular to the given plane, we need to convert the given equation of plane into normal form.

The given equation of the plane isr. 6 i^ - 3 j^ - 2 k^ + 1 = 0r. 6 i^ - 3 j^ - 2 k^  =  -1r. -6 i^ + 3 j^ + 2 k^ = 1 ... 1Now, -62 + 32 + 22 36 + 9 + 4 = 7Dividing (1) by 7, we getr. -67 i^ + 37 j^ + 27 k^ = 17, which is in the normal form  rn d,where the unit vector normal to the given plane, n = -67 i^ + 37 j^ + 27 k^So, its direction cosines are -673727

Page No 28.83:

Question 13:

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.

Answer:

Let M be the foot of the perpendicular of the origin P (0, 0, 0) in the plane 2x-3y+4z-6=0.Then, PM is normal to the plane.  So, the direction ratios of PM are proportional to 2, -3, 4.Since PM passes through P (0, 0, 0) and has direction ratios proportional to 2, -3, 4, the equation of PQ isx-02=y-0-3=z-04=r (say)Let the coordiantes of M be 2r, -3r, 4r.Since M lies in the plane 2x - 3y + 4z - 6 = 0,2 2r - 3 -3r + 4 4r - 6 = 04r + 9r + 16r - 6 = 029r - 6 = 0r = 629Substituting the value of r in the coordinates of M, we getM=2r, -3r, 4r=2 629, -3 629, 4 629=1229, -1829, 2429

Page No 28.83:

Question 14:

Find the length and the foot of perpendicular from the point 1,32,2 to the plane 2x-2y+4z+5=0.          [NCERT EXEMPLAR]

Answer:


Let M be the foot of the perpendicular from P1,32,2 on the plane 2x-2y+4z+5=0.

Then, PM is the normal to the plane. So, its direction ratios are proportional to 2, −2, 4.

Since PM passes through P1,32,2, therefore, its equation is

x-12=y-32-2=z-24=λSay

Let the coordinates of M be 2λ+1,-2λ+32,4λ+2.

Now, M lies on the plane 2x-2y+4z+5=0.

22λ+1-2-2λ+32+44λ+2+5=024λ+12=0λ=-12

So, the coordinates of M are 2×-12+1,-2×-12+32,4×-12+2 or 0,52,0.

Thus, the coordinates of the foot of the perpendicular are 0,52,0.

Now,

PM=1-02+32-522+2-02=1+1+4=6

Thus, the length of the perpendicular from the given point to the plane is 6 units.

Page No 28.83:

Question 15:

Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector 2i^+3j^+4k^ to the plane r.2i^+j^+3k^-26=0. Also find image of P in the plane.

Answer:

Let M be the foot of the perpendicular drawn from the point P(2, 3, 4) in the plane r. 2i^ + j^+ 3 k^-26=0 or 2x+y+3z-26=0.
Then, PM is the normal to the plane. So, the direction ratios of PM are proportional to 2, 1, 3.
Since PM passes through P(2, 3, 4) and has direction ratios proportional to 2, 1, 3, so the equation of PM is
x-22=y-31=z-43=r (say)
Let the coordinates of M be (2r + 2, r + 3, 3r + 4). Since M lies in the plane 2x + y + 3z − 26 = 0,so
22r+2+r+3+33r+4-26=04r+4+r+3+9r+12-26=014r-7=0r=12
Therefore, the coordinates of M are 2r+2, r+3, 3r+4=2×12+2, 12+3, 3×12+4=3, 72, 112
Thus, the position vector of the foot of perpendicular are 3i^+72j^+112k^.
Now,
Length of the perpendicular from P on to the given plane
=2×2+1×3+3×4-26 4+1+9=714=72 units
Let Qx1, y1, z1 be the image of point P in the given plane.
Then, the coordinates of M are x1+22, y1+32, z1+42.
But, the coordinates of M are 3, 72, 112.
x1+22, y1+32, z1+42=3, 72, 112x1+22=3, y1+32=72, z1+42=112x1=4, y1=4, z1=7
Thus, the coordinates of the image of the point P in the given plane are (4, 4, 7).



Page No 28.84:

Question 1:

The plane 2x − (1 + λ) y + 3λz = 0 passes through the intersection of the planes
(a) 2xy = 0 and y − 3z = 0
(b) 2x + 3z = 0 and y = 0
(c) 2xy + 3z = 0 and y − 3z = 0
(d) None of these

Answer:

(a) 2xy = 0 and y − 3z = 0

The given plane is2x - 1 + λ y + 3λz = 02x - y + λ -y + 3z = 0So, this plane passes through the intersection of the planes2x - y = 0 and  -y + 3z = 02x - y = 0 and y - 3z = 0

Page No 28.84:

Question 2:

The acute angle between the planes 2xy + z = 6 and x + y + 2z = 3 is
(a) 45°
(b) 60°
(c) 30°
(d) 75°

Answer:

(b) 60°

We know that the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given bycos θ=a1a2 + b1b2 + c1c2a12 + b12 + c12 a22 + b22 + c22So, the angle between 2x - y + z = 6 and x + y + 2z = 3 is given bySo, cos θ =2 1 + -1 1 + 1 222 + -12 + 12 12 + 12 + 22 = 2 - 1 + 24 + 1 + 1 1 + 1 + 4 = 36 6 = 36 = 12θ = cos-112 = 60o

Page No 28.84:

Question 3:

The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + yz = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is

(a) 7x − 2y + 3z + 81 = 0
(b) 23x + 14y − 9z + 48 = 0
(c) 51x − 15y − 50z + 173 = 0
(d) None of these

Answer:

(c) 51x − 15y − 50z + 173 = 0

The equation of the plane passing through the line of intersection of the given planes isx + 2y + 3z - 4 + λ 2x + y - z + 5 = 0 1 + 2λx + 2 + λy + 3 - λz - 4 + 5λ = 0... 1This plane is perpendicular to 5x + 3y + 6z + 8 = 0. So,5 1 + 2λ + 32 + λ + 6 3 - λ = 0 (Because a1a2 + b1b2 + c1c2 = 0)5 + 10λ + 6 + 3λ + 18 - 6λ = 07λ + 29 = 0λ = - 29   7Substituting this in (1), we get1 + 2 - 297x + 2 + - 297y + 3 + 297z - 4 + 5 - 29  7 = 0 51x +15y - 50z +173 = 0

Page No 28.84:

Question 4:

The distance between the planes 2x + 2yz + 2 = 0 and 4x + 4y − 2z + 5 = 0 is
(a) 12

(b) 14

(c) 16

(d) None of these

Answer:

(c) 16

Multiplying the first equation of the plane by4x+4y-2z+4=0 4x+4y-2z=-4 ... 1The second equation of the plane is4x+4y-2z+5=04x+4y-2z=-5 ... 2We know that the distance between two planes ax+by+cz=d1 and ax+by+cz=d2 is d2-d1a2+b2+c2So, the required distance -5+442+42+-22=-116+16+4=136=16 units



Page No 28.85:

Question 5:

The image of the point (1, 3, 4) in the plane 2xy + z + 3 = 0 is
(a) (3, 5, 2)
(b) (−3, 5, 2)
(c) (3, 5, −2)
(d) (3, −5, 2)

Answer:

(b) (−3, 5, 2)

Let Q be the image of the point P (1, 3, 4) in the plane 2x-y+z+3=0Then PQ is normal to the plane. So, the direction ratios of PQ are proportional to 2, -1, 1.Since PQ passes through P (1, 3, 4) and has the direction ratios proportional to 2, -1, 1., equation of PQ isx-12=y-3-1=z-41=r (say)Let the coordinates of Q be 2r+1, -r+3, r+4. Let R be the mid point of PQ. Then,R=2r+1+12, -r+3+32, r+4+42=r+1, -r+62, r+82Since R lies in the plane 2x-y+z+3=0,2 r+1--r+62+r+82+3=04r+4+r-6+r+8+6=06r+12=0r=-2Substituting this in the coordinates of Q, we getQ=2r+1, -r+3, r+4.=2 -2+1, 2+3, -2+4=-3, 5, 2
So, the answer is (b).

Page No 28.85:

Question 6:

The equation of the plane containing the two lines x-12=y+1-1=z-03 and x-2=y-2-3=z+1-1 is
(a) 8x + y − 5z − 7 = 0
(b) 8x + y + 5z − 7 = 0
(c) 8xy − 5z − 7 = 0
(d) None of these

Answer:

(d) None of these

x12=y+11=z03 and  x2=y23=z+11

Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.

Page No 28.85:

Question 7:

The equation of the plane r=i^-j^+λi^+j^+k^+μi^-2j^+3k^ in scalar product form is
(a) r·5i^-2j^-3k^=7
(b) r·5i^+2j^-3k^=7
(c) r·5i^-2j^+3k^=7
(d) None of these

Answer:

(a) r·5i^-2j^-3k^=7

We know that the equation r=a+λb+μc represents a plane passing through a point whose position vector is a and parallel to the vectors b and c.Here, a = i^- j^ + 0 k^;  b = i^+j^+k^;  c = i^-2 j^+3 k^Normal vector, n=b×c=i^j^k^1111-23=5 i^-2 j^-3 k^The vector equation of the plane in scalar product form isr. n=a. nr. 5 i^-2 j^-3 k^=i^ - j^ + 0 k^. 5 i^ - 2 j^ - 3 k^r. 5 i^ - 2 j^ - 3 k^ = 5 + 2 + 0r. 5 i^ - 2 j^ - 3 k^ = 7r. 5 i^ - 2 j^ - 3 k^ = 7

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Question 8:

The distance of the line r=2i^-2j^+3k^+λi^-j^+4k^ from the plane r·i^+5j^+k^=5 is
(a) 533

(b) 1033

(c) 2533

(d) None of these

Answer:

(b) 1033

The given line passes through the point whose position vector is a=2 i^-2 j^+3 k^.We know that the perpendicular distance of a point P of position vector a from the plane rn = d is given by p=a. n-dnHere,a=i^-2 j^+3 k^; n=i^+5 j^+k^; d=5So, the required distance p is given byp=i^-2 j^+3 k^. i^+5 j^+k^-5i^+5 j^+k^=2-10+3-51+25+1=-1027=103 3 units

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Question 9:

The equation of the plane through the line x + y + z + 3 = 0 = 2xy + 3z + 1 and parallel to the line x1=y2=z3 is
(a) x − 5y + 3z = 7
(b) x − 5y + 3z = −7
(c) x + 5y + 3z = 7
(d) x + 5y + 3z = −7

Answer:


(a) x − 5y + 3z = 7

The equation of the plane passing through the line of intersection of the given planes isx + y + z + 3 + λ 2x - y + 3z + 1 = 0 1 + 2λx + 1 - λy + 1+ 3λz + 3 + λ = 0... 1This plane is parallel to the line x1 = y2 = z3. It means that this line is perpendicular to the normal of the plane (1).1 1 + 2λ + 2 1 -λ + 3 1 + 3λ=0  (Because a1a2 + b1b2 + c1c2 = 0)1 + 2λ + 2 - 2λ + 3 + 9λ = 09λ+6=0λ=-23Substituting this in (1), we get1+2 -23x+1--23y+1+3 -23z+3+-23=0-x+5y-3z+7=0x-5y+3z=7

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Question 10:

The vector equation of the plane containing the line r=-2i^-3j^+4k^+λ3i^-2j^-k^ and the point i^+2j^+3k^ is

(a) r·i^+3k^=10

(b) r·i^-3k^=10

(c) r·3i^+k^=10

(d) None of these

Answer:


(a) r·i^+3k^=10

Let the direction ratios of the required plane be proportional to a, b, c.Since the required plane contains the line r= -2 i^-3 j^+4 k^+λ 3 i^-2 j^-k^, it must pass through the point (-2, -3, 4) and it should be parallel to the line.So, the equation of the plane is a x + 2 + b y + 3 + c z - 4 = 0 ... 1 and3a - 2b - c = 0 ... 2It is given that plane (1) passes through the point i^ + 2 j^ + 3 k^ or (1, 2, 3).  So,a 1 + 2 + b 2 + 3 + c 3 - 4 = 0 3a+5b-c=0 ... 3Solving (1), (2) and (3), we getx+2y+3z-43-2-135-1=07 x+2+0 y+3+21 z-4=0x + 2 + 3z - 12=0x + 3z = 10 or ri^ +3 k^ = 10

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Question 11:

A plane meets the coordinate axes at A, B and C such that the centroid of ∆ABC is the point (a, b, c). If the equation of the plane is xa+yb+zc=k, then k =
(a) 1
(b) 2
(c) 3
(d) None of these

Answer:

(c) 3

Let α, β and γ be the intercepts of the given plane on the coordinate axes.Then, the plane meets the coordinate axes atA α, 0, 0, B 0, β, 0 and C=0, 0, γGiven that the centroid of the triangle = a, b, cα + 0 + 03, 0 + β + 03, 0 + 0 + γ3 = a, b, cα3, β3, γ3 = a, b, cα3 = aβ3 = bγ3 = cα = 3a, β = 3b, γ = 3c ... 1Equation of the plane whose intercepts on the coordinate axes are αβ and γ isxα + yβ + zγ = 1x3a + y3b + z3c = 1 [From (1)]xa + yb + zc = 3

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Question 12:

The distance between the point (3, 4, 5) and the point where the line x-31=y-42=z-52 meets the plane x + y + z = 17 is
(a) 1
(b) 2
(c) 3
(d) None of these

Answer:

(c) 3

The coordinates of any point on the given line are of the formx-31=y-42=z-52=λx = λ + 3;  y = 2λ + 4;  z = 2λ + 5So, the coordiantes of the point on the given line are λ+3, 2λ+4, 2λ+5. This point lies on the plane,     x + y + z = 17λ + 3 + 2λ + 4 + 2λ + 5 = 17                                                               5λ = 5λ = 1So, the coordinates of the point areλ+3, 2λ+4, 2λ+5=1+3, 2 1+4, 2 1+5=4, 6, 7Now, the distance between the points 4, 6, 7 and 3, 4, 5 is3-42+4-62+5-72=1+4+4=3 units
So, the answer is (c).

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Question 13:

A vector parallel to the line of intersection of the planes r·3i^-j^+k^=1 and r·i^+4j^-2k^=2 is
(a) -2i^+7j^+13k^

(b) 2i^+7j^-13k^

(c) -2i^-7j^+13k^

(d) 2i^+7j^+13k^

Answer:

(b) 2i^+7j^-13k^

Let the required vector be a i^ + b j^ + c k^ ... 1Since the vector is parallel to the line of intersection of the given planes, 3a - b + c = 0 ... 2a + 4b - 2c = 0 ... 3Solving (2) and (3), we get  a-2 = b7 = c13Substituting these values in (1), we get-2 i^ + 7 j^ + 13 k^, which is the required vector. 

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Question 14:

If a plane passes through the point (1, 1, 1) and is perpendicular to the line x-13=y-10=z-14 then its perpendicular distance from the origin is
(a) 3/4
(b) 4/3
(c) 7/5
(d) 1

Answer:

(c) 7/5

Since the plane is perpendicular to the given line, its direction ratios are proportional to 3, 0, 4So, the required equation of the plane is of the form3x + 0y + 4z + d = 0... 1, where d is a constant.Since this plane passes through (1, 1, 1),3 + 0 + 4 + d = 0d =-7Substituting this in (1), we get3x + 0y + 4z - 7 = 0 ... 2Perpendicular distance of (2) from the origin=3 0 + 0 + 4 0 - 732 + 02 + 42=0 + 0 - 725=75 units

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Question 15:

The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is
(a) x − 4y + 2z + 4 = 0
(b) x + 4y + 2z + 4 = 0
(c) x − 4y + 2z − 4 = 0
(d) None of these

Answer:

a) x − 4y + 2z + 4 = 0

Let a, b, c be the direction ratios of the required plane.The given line equations can be rewritten asx-11 = y-5212 = z-012 ... 1x-0 13 = y-11414 = z-4313 ... 2Since the required plane is parallel to the lines (1) and (2),a + b2 + c2 = 0 2a + b + c = 0 ... 3a3 + b4 + c3 = 0 4a + 3b + 4c = 0 ... 4Solving (3) and (4) using cross-multiplication method, we geta1 = b-4 = c2 = λ (say)a = λ, b = -4λ, c = 2λNow, the equation of the plane whose direction ratios are λ, -4λ, 2λ and passing through the point (2, 3, 3) isλ x - 2 + -4λy - 3 + 2λ z - 3 = 0x - 4y + 2z + 4 = 0
So, the answer is (a).



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Question 16:

The distance of the point (−1, −5, −10) from the point of intersection of the line r = 2i^ - j ^+ 2k^ + λ3i^ + 4j^ + 12k^ and the plane r·i^ - j ^+ k^ = 5 is
(a) 9
(b) 13
(c) 17
(d) None of these

Answer:

(b) 13

Given equation of line isr = 2 i^ - j ^+ 2 k^ + λ 3 i^ + 4 j ^+ 12 k^r = 2 + 3λ i^ + -1 + 4λ j ^+ 2 + 12λ k^The coordinates of any point on this line are of the form 2 + 3λ i^ + -1 + 4λ j ^+ 2 + 12λ k^. or 2 + 3λ, -1 + 4λ, 2 + 12λSince this point lies on the plane ri^ - j^ + k^ = 5 ,2 + 3λ i^ + -1 + 4λ j^ + 2 + 12λ  k^. i^ - j^ + k^  = 5 2 + 3λ + 1 - 4λ + 2 + 12λ - 5 = 0λ = 0So, the coordinates of the point are    2 + 3λ, -1 + 4λ, 2 + 2λ=2 + 0, -1 + 0,  2 + 0=2, -1, 2Distance between (2, -1, 2) and (-1, -5, -10)=-1 -22 + -5 + 12 + -10 - 22=9 +16+144=13 units

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Question 17:

The equation of the plane through the intersection of the planes ax + by + cz + d = 0 and lx + my + nz + p = 0 and parallel to the line y=0, z=0
(a) (blam) y + (clan) z + dlap = 0
(b) (ambl) x + (mcbn) z + mdbp = 0
(c) (nacl) x + (bncm) y + ndcp = 0
(d) None of these 

Answer:

The equation of the plane passing through the intersection of the planes
ax + by + cz + d = 0
and lx + my + nz + p = 0
will be (​ax + by + cz + d) + λ(​lx + my + nz + p) = 0

x(a + ​λl) + y(b + ​λm) + z(c + ​λn) + (d + ​λp)=0  .......(1)

Since the plane is parallel to the line y=0 and z=0
a + ​λl=0
λ = -al
putting the value of ​λ in equation (1), we get
x(a + (-al)l) + y(b + (-al)m) + z(c + (-al) n) + d + (-al)p =0

 y(bl - am) + z(cl -an) + dl - ap =0

Hence, option (a)

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Question 18:

The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is
(a) x + y + z = 1
(b) x + y + z = 0
(c) x + yz = 1
(d) x + y + z = 2

Answer:

(a) x + y + z = 1

We know that the equation of aplane whose intercepts are a, b, c isxa+yb+zc=1 ... 1It is given that a = b = cSo, from (1),xa + ya + zc = 1 x + y + z = a ... (2)Since it is given that the intercepts of the required plane are of unit length,a = b = c = 1Substituting a = 1 in (2), we getx + y + z = 1

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Question 19:

The locus represented by xy + yz = 0 is
(a) a pair of perpendicular lines
(b) a pair of parallel lines
(c) a pair of parallel planes
(d) a pair of perpendicular planes

Answer:

Given xy + yz = 0
i.e. (x + z)y = 0
i.e. x + z = 0 or y = 0
Which represent two plane 
For plane y = 0, normal is j^ and normal to plane x + z = 0 is i^+k^
Since (i^+k^).j^ =0
⇒ Two planes are perpendicular        (∵ normals are perpendicular) 

Hence, the correct answer is option D. 

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Question 20:

The reflection of the point (α, β, γ) in the xy-plane is
(a) (α, β, 0)                 
(b) (0, 0, γ)                 
(c) (-α, -β, γ)               
(d) (α, β, -γ)

Answer:

Reflection in xy plane, changes the z- co-ordinate by negative (i.e. an other side of z-axis) 

∴ Reflection of α, β, γ in the xy plane is α, β,-γ 

Hence, the correct answer is option D.

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Question 21:

The plane 2x-3y+6z-11=0 makes an angle sin-1α with x-axis. The value of α is equal to
(a) 32

(b) 23

(c) 27

(d)  37

Answer:

Given, plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1α with x-axis
Since n=2i^-3j^+6k^ for given plane and sinθ=b.nbnHere θ is angle between n and b where b=i^ x-axis is given sinsin-1α=i^.2i^-3j^+6k^12 22 +32+62i.e. sinsin-1α=214+36+9i.e. α=27
Hence, the correct answer is option C.

 

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Question 22:

The sine of the angle between the straight line x-23=y-34=z-45 and the plane 2x-2y+z=5 is 
(a) 1065

(b) 452

(c) 235

(d) 210

Answer:

For given line, x-23=y-34=z-45

Direction ratios are given by (3, 4, 5) = (a, b, c) say and for plane, 2x − 2y + z = 5,
Normal vector is given by (2, −2, 1) = (A, B, C
∴ Since of angle between the straight line and plane is 
sinθ=aA+bB+cCa2+b2+c2 A2+B2+C2      =3×2+4×-2+5×132+42+52 22+22+1=6-8+550 9=3152i.e. sinθ=315×2=152=1×252×2=25×2i.e. sinθ=210

Hence, the correct answer is option D.

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Question 23:

The distance of the plane r.27i^+37j^-67k^=1 from the origin is 
(a) 1

(b) 7

(c) 17

(d) none of these

Answer:

For given plane,

r.27i^+37j^-67k^=1Let r=xi^+yj^+zk^i.e. xi^+yj^+zk^.27i^+37j^-67k^=1i.e. 27x+37y-127z=1

Since for equation of plane ax + by + cz = d, d represents distance from origin.

Hence, for 27x+37y-67z=1

Hence, the correct answer is option A.

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Question 1:

If the plane x+2y-2z=d, d>0 is at a 5 unit distance from the point (1, -2, 1),then d = ____________.

Answer:

For given plane, x + 2y – 2z = d
Distance of this plane from (1, –2, 1) is 5 is given
i.e. 5=1+2-2-21-d12+22+22i.e. 5=1-4-2-d9i.e. 15=-5-di.e. 15=5+di.e. d=10
 

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Question 2:

The planes 3x-6y-2z=7 and 2x+y-λz=5 are perpendicular. Then the value of λ is __________.

Answer:

Given, planes, 3x-6y-2z=7 and 2x+y-λz=5 are perpendicular

Since two planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 are perpendicular If a1 a2 +b1 b2 +c1 c2=0i.e. 32+-61+-2-λ=0i.e. 6-6+2λ=0i.e. 2λ=0i.e. λ=0

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Question 3:

The equation of the plane passing through (1, 2, 3) and parallel to the plane 2x+3y-4z=0 is __________.

Answer:

For plane normal is given by (2, 3, −4)            [∴being parallel to the plane 2x + 3y − 4z = 0] 

i.e. n=2, 3,-4 and α=1, 2, 3    given that plane passes through 1, 2, 3 For r=xi^+yj^+2k^,Equation of plane is r-α.n=0i.e. r.n=α.ni.e. xi^+yj^+zk^,2i^+3j^-4k^=i^+2j^+3k^.2i^+3j^-4k^i.e. 2x+3y-4z=2+6-12i.e. 2x+3y-4z=-4
i.e. 2x + 3y – 4z + 4 = 0 is the equation of plane which passes through (1, 2, 3) and is parallel to 2x + 3y – 4z = 0



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Question 4:

The equation of a line passing through (1, 2, 3) and normal to the plane 2x-3y+6z=11 is _______________.

Answer:

Since, line passes through (1, 2, 3)
i.e. (x1, y1, z1) (1, 2, 3) and normal is to the plane 2x − 3y + 6z = 11
∴ Direction co-ordinates are (2, −3, 6) = (a, b, c
Since equation of line passing through (x1, y1, z1) and in the direction (a, b, c) is 
x-x1a=y-y1b=z-z1c equation of line is given by x-12=y-2-3=z-36

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Question 5:

The sum of the intercepts of the plane 2x+3y-4kz=24 on the coordinates is 8, then the value of k is ____________.

Answer:

x intercept of the plane 2x + 3y − 4kz = 24 is obtained where y = 0 = z
i.e. 2x = 24 
i.e. x = 12
x intercept is 12. 
Similarly y-intercept is obtained where x = 0 = z 
i.e. 3y = 24 
i.e. y = 8 is the y - intercept 
and z-intercept is obtained when x = 0 = y 
i.e. −4kz =24
 i.e. z=-6k is the z-intercept 
Since sum of all intercepts is 8 (given)
i.e. 12+8+-6k=8i.e. 12=6ki.e. k=12

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Question 6:

If the distance of the point (1, 1, 1) from the origin is half its distance from the plane x + y + z + k = 0, then the values of k are ___________.

Answer:

Given, distance of the point (1, 1, 1) from the origin is the half its distance from the plane x + y + z + k = 0

Since distance between (1, 1, 1) and (0, 0, 0) =1-02+1-02+1-02=3       ....1

Distance of (1, 1, 1) from (x + y + z + k = 0)

d=1+1+1+k1+1+1=3+k3      ....2Since 1=12 2 is given 3=123+k3i.e. 2×3=3+ki.e. ±6=3+ki.e. k=3 or k=-9

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Question 7:

If the plane xa+yb+zc=3 meets the coordinates axes at A, B and C, then the coordinates of the centroid of ∆ABC are ______________.

Answer:

Given plane, xa+yb+zc=3

Meets x-axis when y = 0 = zxa=3

i.e. x = 3a

Plane meets y-axis when x = 0 = z,

i.e. yb=3i.e. y=3b

and this plane meets z-axis when x = 0 = y

i.e. zc=3i.e. z=3c

∴ A is given by (3a, 0, 0)

B is given by (0, 3b, 0)

C is given by (0, 0, 3c)

 Centroid is 3a+0+03,0+3b+03,0+0+3c3

i.e. Centroid of ΔABC = (a,b,c)

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Question 8:

If O is the origin, then the equation of the plane passing through P(a, b, c) and perpendicular to OP is _____________.

Answer:

Since plane is perpendicular to OP and passes through (a, b, c)

Equation of plane is given by a(x − a) + b(y − b) + c(z − c) = 0 

i.e. ax − a2 + by − b2 + cz − c2 = 0

i.e. ax + by + cz = a2 + b2 + c2

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Question 9:

A variable plane moves so that the sum of the reciprocals of its intercepts on the coordinate axis is 12. The plane passes through the point ____________.

Answer:

For a variable plane, given sum of reciprocal of its intercept is 12 let a,b,c be the intercepts with co-ordinates axes according to given condition, 
1a+1b+1c=12i.e. 2a+2b+2c=1i.e. 21a+21b+21c=1
i.e plane passes through fixed point (2, 2, 2,)

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Question 10:

The value of α for which the plane x + αy + z = 5 cuts equal intercepts on the axes, is ____________.

Answer:

Equation of a plane which equal intercept on the axes is xa+4a+za=1
Since x + αy + z = 5 cuts equal intercept

i.e. x5+α5y+z5=1i.e. 15=α5i.e. α=1

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Question 11:

If the line 2x-12=2-y2=z+1k is parallel to the plane 2x-y+z=3, then k = _______________.

Answer:

 Given line2x-12=2-y2=z+1k is parallel to plane 2x-y+z=3
Line is of the pr  x-121=y-2-2=z--1k

is parallel if n for plane is perpendicular to line for plane 2x – y + z = 3

Normal is n 2i^-j^+k^ and b=i^-2j^+kk^Since n.b=0i.e. 21-1-2+1k=0i.e. 2+2+k=0i.e. k=-4
 

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Question 12:

If the line x-23=y-1-5=z+22 lies in the plane x+3y-αz+7=0, then α = _________________.

Answer:

Given line x-23=y-1-5=z+22
Lies in the plane x + 3yαz + 7 = 0
Whose normal n=i^+3j^-αk^ then b=3i^-5j^+2k^n=i^+3j^-αk^i.e. 31-53+2-α=0i.e. 3-15-2α=0i.e. -12-2α=0i.e. 2α=-12i.e. α=-6

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Question 13:

If line x+1λ=y-11=z+2-4 is perpendicular to the plane 2x+2y-8z+5=0. Then the value of λ is _____________.

Answer:

Given line, x+1λ=y-11=z+2-4 is perpendicular to the plane 2x+2y-8z+5=0

Normal n for the plane is n=2i^+2j^-8k^ and direction vector for line b=λi^+j^-4k^

Since line is perpendicular to plane

i.e. 2λ=21=-8-4i.e. nand b have proportional vectors i.e. 2λ=21i.e. λ=1

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Question 14:

A plane passes through the points (2, 0, 0), (0, 3, 0) and (0, 0, 4). The equation of the plane is _______________.

Answer:

Suppose equation of plane is a( x1) + b(y − y1) + c(z − z1) = 0
Where plane passes through (x1, y1, z1)
Given plane passes through (2, 0, 0), (0, 3, 0) and (0, 0, 4) 
i.e. a(x − 2) + b(y − 0) + c(z − 0) = 0    ...(1) 

Since equation (1) passes through (0, 3, 0) and (0, 0, 4) 
We get, a(0 − 2) + b(3 − 0) + c(0 − 0) = 0
i.e. −2a + 3b = 0 and a(0 − 2) + b(0 − 0) + c(4 − 0) = 0
i.e.  −2a + 4= 0
i.e. a = 2c and 3b = 2a = 4c
∴ equation of plane is 2c(x − 2) + 43c (− 0) + c(z − 0) = 0 
2(x − 2) + 43y + z = 0
i.e. 2x43y + z − 4 × 3 = 0
i.e. 6x + 4y + 3z − 12 = 0 is the equation of the plane

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Question 15:

The cartesian equation of the plane r·( i^+j^+k^ )=2 is _____________.

Answer:

The cartesian equation of the plane 

r.i^+j^+k^=2for r=xi^+yj^+zk^is xi^+yj^+zk^.i^+j^+k^=2i.e. x+y+z=2 is the cartesian equation 

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Question 16:

The intercepts made by the plane 2x-3y+5z+4=0 on the coordinate axes are _____________.

Answer:

For given plane 2x − 3y + 5z = −4
Intercept on x -axis will be given by y = 0 and z = 0 
i.e. 2x = −4 
i.e. x = (2) 
i.e. x = −2 is the x-intercept 
Intercept on y-axis will be given by x = 0 and z = 0 
i.e. −3y = −4 
i.e. y=43 is the y-intercept 
and for z-intercept will be given by x = 0 and y = 0 
i.e. 5z = −4 
i.e. z=-45 is the z-intercept 
∴ -2, 43 and -45 are the intercept made by the plane 2x − 3y + 5z + 4 = 0 on the co-ordinate axes.

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Question 17:

The angle between the line r=(5i^-j^-4k^ )+λ(2i^-j^+k^ ) and the plane r·(3i^-4j^-k^ )+5=0 is ____________.

Answer:

Since the angle between the line 

r=a+λb and plane r.n=d is given by sinθ=b.nbnGiven line r=5i^-j^-4k^ +λ2i^-j^+k^b=2i^-j^+k^ and the plane r.3i^-4j^-k^+5=0n=3i^-4j^-k^
i.e. sinθ=2i^-j^+k^.3i^-4j^-k^2i^-j^+k^3i^-4j^-k^=6+4-14+1+1 9+16+1=96 26 sinθ =9156
i.e. angle between given line and plane is sin-1 9156

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Question 18:

The foot of the perpendicular from the origin to a plane has the coordinates (5,-3,-2). The equation of the plane is ____________. 

Answer:

Given foot of perpendicular from the origin to a plane has the co-ordinates (5, −3, −2)
i.e. (5, −3, −2) lie an plane. 
i.e. equation of plane is a(x − 5) + b(y + 3) + c( z + 2) = 0
Since components of normal are = 5-0i^+-3-0j^+-2-0k^
i.e. a = 5, b = −3, c = −2 
i.e. 5(x − 5) + (−3) (y + 3) + (−2) (z + 2) = 0
i.e. 5x − 25 − 3y − 9 − 2− 4 = 0
i.e. 5x − 3− 2z – 38 = 0
i.e. 5x − 3− 2z = 38
i.e. for r=xi^+4j^+zk^r.5i^3j^2k^=38 is the required eqaution of plane 



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Question 1:

Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5).

Answer:

The equation of the plane parallel to the plane XOY is z=b ... 1, where b is a constant.It is given that this plane passes through (2,-3, 5). So,5=bSubstituting this value in (1), we get z=5, which is the required equation of the plane.

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Question 2:

Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).

Answer:

The equation of the plane parallel to the plane YOZ  is x = b ... 1, where b is a constant.It is given that this plane passes through (-4, 1, 0). So,-4 = bSubstituting this value in (1), we get x=-4 , which is the required equation of the plane.

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Question 3:

Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).

Answer:

The equation of the plane passing through (a, 0, 0), (0, b, 0) and (0, 0, c) is
x-ay-0z-00-ab-00-00-a0-0c-0 = 0 x-ay z-ab0-a0c =0bc x-a + acy + abz = 0bcx + acy + abz = abcDividing the equation by abc, we getxa + yb + zc = 1

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Question 4:

Write the general equation of a plane parallel to X-axis.

Answer:

The general equation of a plane isax + by + cz + d = 0 ... 1This plane is parallel to the x - axis.It means that this plane passes through the point 0, y, z. So,a 0 + by + cz + d = 0by + cz + d = 0

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Question 5:

Write the value of k for which the planes x − 2y + kz = 4 and 2x + 5yz = 9 are perpendicular.

Answer:

We know that the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are perpendicular to each other only if a1a2 + b1b2 + c1c2 = 0The given planes are x - 2y + kz = 4 and 2x + 5y - z = 9a1 = 1; b1 = -2; c1 = k; a2 = 2; b2 = 5; c2 = -1It is given that the given planes are perpendicular. a1a2 + b1b2 + c1c2 = 01 2 + -2 5 + k -1 = 02 - 10 - k = 0-8 - k = 0k = -8

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Question 6:

Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.

Answer:

The given equation of the plane is2x - 3y + 4z = 12Dividng both sides by 12, we get2x12 + -3y12 + 4z12 = 1212x6 + y-4 + z3 = 1 ... 1We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c isxa + yb + zc = 1 ... 2Comparing (1) and (2), we geta= 6,  b =-4 and c = 3

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Question 7:

Write the ratio in which the plane 4x + 5y − 3z = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2).

Answer:

We know that the ratio in which the plane ax+by+cz+d=0 divides the line segment joining x1, y1, z1 and x2, y2, z2  is- ax1+by1+cz1+dax2+by2+cz2+dHere, a = 4; b = 5; c = -3;  d = -8; x1 = -2; y1 = 1;  z1 = 5;  x2 = 3;  y2 = 3;  z2 = 2So, the required ratio=- 4 -2+5 1-3 5-84 3+5 3-3 2-8=- -8+5-15-812+15-6-8=2613=21 or 2 : 1

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Question 8:

Write the distance between the parallel planes 2xy + 3z = 4 and 2xy + 3z = 18.

Answer:

The given equations are 2x - y + 3z = 4 ... 1The second equation of the plane is2x - y + 3z = 18 ... 2We know that the distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is d2-d1a2+b2+c2So, the required distance  is 18-422+-12+32=144+1+9=1414=14 units

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Question 9:

Write the plane r·2i^+3j^-6k^=14 in normal form.

Answer:

The given equation of the plane isr. 2 i^+3 j^-6 k^ = 14 or r. n =14, where n =  2 i^ + 3 j^ - 6 k^n=4+9+36=7For reducing the given equation to normal form, we need to divide it by n. Then, we getr. nn=14nr. 2 i ^+ 3 j^ - 6 k^7 = 147r. 27 i^ + 37 j^ - 67k^ = 2, which is the required normal form.

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Question 10:

Write the distance of the plane r·2i^-j^+2k^=12 from the origin.

Answer:

The given equation of the plane isr. 2 i^- j^+2 k^ = 12  or  r. n = -6,  where n =  2 i ^-  j^ + 2 k^n = 4+1+4 = 3For reducing the given equation to normal form, we need to divide both sides by n. Then, we getr. nn=12nr. 2 i^- j^+2 k^3 = 123r. 23 i^-13 j^+23k^ = 4 ... 1The equation of the plane in normal form isr. n^ = d ... 2(where is the distance of the plane from the origin)Comparing (1) and (2),length of the perpendicular from the origin to the plane = d = 4 units

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Question 11:

Write the equation of the plane r=a+λb+μc in scalar product form.

Answer:

The equation of the given plane isr= a + λ b + μ cSo, the plane passes through the vector a and parallel to the vectors b and c.So, the plane passes through the vector a whose normal vector is b × a (It means that n=b×a)So, the equation of the plane in scalar product form isr-a. n=0r-a. b×c=0

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Question 12:

Write a vector normal to the plane r=lb+mc.

Answer:

The equation of the given plane isr=l b+m cSo, the plane passes parallel to the vectors b and c.So, the vector normal to the plane is b × c.

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Question 13:

Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2yz = 7.

Answer:

Let the equation of a plane parallel to the given plane be 3x+2y-z=k ... 1This passes through (2, -1, 1). So, 3 2+2 -1-1=kk=3Substituting this in (1), we get3x+2y-z=3, which is the equation of the required plane.

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Question 14:

Write the equation of the plane containing the lines r=a+λb and r=a+μc.

Answer:

The given plane passes through the linesr=a+λ b and r=a+μ cSo, the plane passes through the vector a and parallel to the vectors b and c.So, the plane passes through the vector a whose normal vector is b × a. (It means that n=b×a)So, the equation of plane in scalar product form isr-a. n=0r-a. b×c=0

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Question 15:

Write the position vector of the point where the line r=a+λb meets the plane r.n=0.

Answer:

Given equation of the line isr =a  + λ b... 1Given equation of the plane isr. n = 0a + λ b. n = 0 [From (1)]a. n + λ b. n = 0λ = -a. nb. nSubstituting this in (1), we getr=a-a. nb. n b, which is the required position vector that lies both on the line and the plane.

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Question 16:

Write the value of k for which the line x-12=y-13=z-1k is perpendicular to the normal to the plane r·2i^+3j^+4k^=4.

Answer:

Direction ratios of the given line x-12 = y-13 = z-1k are proportional to 2, 3, k.Direction ratios of the normal to the plane r2 i ^+ 3 j^ + 4 k^ = 4 are 2, 3, 4.Given that these two are perpendicular.2 2+3 3 + k4 = 0 (Because a1a2 + b1b2 + c1c2 = 0)4 + 9 + 4k =013 + 4k = 0k = -134

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Question 17:

Write the angle between the line x-12=y-21=z+3-2 and the plane x + y + 4 = 0.

Answer:

The given line is parallel to the vector b=i^ + 2 j^ + 2 k^ and the given plane is normal to the vector n=i^ + j^+0 k^.We know that the angle θ between the line and the plane  is given bysin θ=b. nb n=i^ + 2 j^ + 2 k^. i ^+ j ^+ 0 k^i^ + 2 j^ + 2 k^ i^ + j^ + 0 k^=1 + 2 + 01 + 4 + 4 1 + 1 + 0 = 33 2 = 12θ = sin-1 12 = 45o

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Question 18:

Write the intercept cut off by the plane 2x + yz = 5 on x-axis.

Answer:

For x- interecept, put = 0 and = 0 in the given equation. Then, we get2x+0-0=52x=5x=52

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Question 19:

Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.

Answer:

We know that the distance of the point x1, y1, z1 from the plane ax + by + cz + d = 0 is given byax1 + by1 + cz1 + da2 + b2 + c2So, the required distance =2 0-3 0+6 0+2122 + -32 + 62=214 + 9 + 36=217=3  units

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Question 20:

Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane r·2i^+j^+2k^=5.

Answer:

The required line is normal to the plane r.2 i^+j^+2 k^=5 and it is parallel to the normal vector of the plane.So, the required line is parallel to the vector b=2 i^+j^+2 k^It is given that the line passes through the point (1, -2, -3) whose position vector is given by a=i^-2 j^-3 k^.We know that the equation of the line passing through the point whose position vector is a and parallel to the vector b is given byr=a+λ br=i^ - 2 j^ - 3 k^ + λ 2 i^ + j^ + 2 k^

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Question 21:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question:

Find the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane r.i^+j^+k^=2.             [CBSE 2014]

Answer:


The required plane passes through ai^+bj^+ck^ and is parallel to the plane r.i^+j^+k^=2.

So, it is normal to the vector i^+j^+k^ which is normal to the given plane.

Hence, the vector equation of the required plane is

r-ai^+bj^+ck^.i^+j^+k^=0                                  r-a.n=0r.i^+j^+k^=ai^+bj^+ck^.i^+j^+k^r.i^+j^+k^=a+b+c

Thus, the vector equation of the required plane is r.i^+j^+k^=a+b+c.

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Question 22:

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is 2i^-3j^+6k^.

Answer:

Given: 
Normal vector, n^ = 2i ^- 3j ^+ 6k^Perpendicular distance, d = 5 units

The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector  n^ = 2i ^- 3j ^+ 6k^ is as follows:

r . n^ = d
r . (2i ^- 3j^ + 6k^) = 5

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Question 23:

Write the equation of a plane which is at a distance of 53 units from origin and the normal to which is equally inclined to coordinate axes.

Answer:

Let α, β and γ be the angles made by n with x, y and z-axes, respectively.It is given thatα=β=γcosα=cosβ=cosγl=m=n, where l,m, n are direction cosines of n.But l2+m2+n2=1l2+l2+l2=13 l2=1l2=13l=13So, l=m=n=13It is given that the length of the perpendicular of the plane from the origin, = 5 3The normal form of the plane is lx+my+nz=p13x+13y+13z=53x+y+z=53 3 x+y+z=15

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Question 24:

Find the acute angle between the planes r·i^-2j^-2k^ =1 and r·3i^-6j^+2k^ = 0.

Answer:

For given plane
r.i^-2j^-2k^=1 and r.3i^-6j^+2k^=0Using formula for plane r.n1=d1 and r.n2=d2cosθ=n1.n2n1n2 where θ is angle between two given planesi.e. cosθ=i^-2j^-2k^.3i^-6j^+2k^i^-2j^-2k^3i^-6j^+2k^i.e. cosθ=3+12-41+4+4 9+36+4i.e. cosθ=119 49=113×7i.e. cosθ=1121i.e. θ=cos-11121



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